# 48 circle part 1 of 2

26 de Oct de 2013
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### 48 circle part 1 of 2

• 1. FREE Download Study Package from website: www.tekoclasses.com Subject : Mathematics Topic : CIRCLE R Index 1. Theory 2. Short Revision 3. Exercise (Ex. 1 to 5) 4. Assertion & Reason 5. Que. from Compt. Exams 6. 34 Yrs. Que. from IIT-JEE 7. 10 Yrs. Que. from AIEEE Student’s Name :______________________ Class :______________________ Roll No. :______________________ ADDRESS: R-1, Opp. Raiway Track, New Corner Glass Building, Zone-2, M.P NAGAR, Bhopal . : (0755) 32 00 000, 98930 58881, www.tekoclasses.com Page : 1 of 30 , CIRCLE STUDY PACKAGE 98930 58881 , BHOPAL, (M.P.) jfpr% ekuo /keZ iz.ksrk ln~xq# Jh j.kNksM+nklth egkjkt TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjr e/;e eu dj ';keA a iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA
• 3. 2. 3. Ans. x 2 + y2 ± 6 2 x – 6y + 9 = 0 Find the equation of a circle which touches y-axis at a distance of 2 units from the origin and cuts an intercept of 3 units with the positive direction of x-axis. Ans. x 2 + y2 ± 5x – 4y + 4 = 0 Parametric Equations of a Circle: FREE Download Study Package from website: www.tekoclasses.com The parametric equations of (x − h) 2 + (y − k) 2 = r2 are: x = h + r cos θ ; y = k + r sin θ ; − π < θ ≤ π where (h, k) is the centre, r is the radius & θ is a parameter. Example : Find the parametric equations of the circle x 2 + y2 – 4x – 2y + 1 = 0 Solution : We have : x 2 + y2 – 4x – 2y + 1 = 0 ⇒ (x 2 – 4x ) + (y2 – 2y) = – 1 ⇒ (x – 2)2 + (y – 1)2 = 22 So, the parametric equations of this circle are x = 2 + 2 cos θ , y = 1 + 2 sin θ. Example : Find the equations of the following curves in cartesian form. Also, find the centre and radius of the circle x = a + c cos θ, y = b + c sin θ x −a y−b Solution : We have : x = a + c cos θ, y = b + c sin θ ⇒ cos θ = , sin θ = c c 2 2 x−a  y −b   +   = cos2θ + sin2θ ⇒ ⇒ (x – a)2 + (y – b)2 = c2  c   c  Clearly, it is a circle with centre at (a, b) and radius c. Self Practice Problems : 1. Find the parametric equations of circle x 2 + y2 – 6x + 4y – 12 = 0 Ans. x = 3 + 5 cos θ, y = –2 + 5 sin θ 2. Find the cartesian equations of the curve x = –2 + 3 cos θ, y = 3 + 3 sin θ Ans. (x + 2)2 + (y – 3) 2 = 9 4. Position of a point with respect to a circle: The point (x 1, y1) is inside, on or outside the circle S ≡ x 2 + y2 + 2gx + 2fy + c = 0. according as S1 ≡ x 1² + y1² + 2gx 1 + 2fy1 + c < , = or > 0. NOTE : The greatest & the least distance of a point A from a circle with centre C & radius r is AC+r& AC − r respectively. Di scuss the posi t ion of t he poi nts (1, 2) and (6, 0) wi th respect to the ci rcl e x 2 + y2 – 4x + 2y – 11 = 0 Solution : We have x 2 + y2 – 4x + 2y – 11 = 0 or S = 0, where S = x 2 + y2 – 4x + 2y – 11. For the point (1, 2), we have S1 = 12 + 22 – 4 × 1 +2 × 2 – 11 < 0 For the point (6, 0), we have S2 = 62 + 02 – 4 × 6 +2 × 0 – 11 > 0 Hence, the point (1, 2) lies inside the circle and the point (6, 0) lies outside the circle. Self Practice Problem : 1. How are the points (0, 1) (3, 1) and (1, 3) situated with respect to the circle x 2 + y2 – 2x – 4y + 3 = 0? Ans. (0, 1) lies on the circle ; (3, 1) lies outside the circle ; (1, 3) lies inside the circle. Example : 5. Line and a Circle: Let L = 0 be a line & S = 0 be a circle. If r is the radius of the circle & p is the length of the perpendicular from the centre on the line, then: (i) p>r ⇔ the line does not meet the circle i. e. passes out side the circle. (ii) p=r ⇔ the line touches the circle. (It is tangent to the circle) (iii) p<r ⇔ the line is a secant of the circle. (iv) p=0 ⇒ the line is a diameter of the circle. Also, if y = mx + c is line and x 2 + y2 = a2 is circle then (i) c2 > a2 (1 + m 2) ⇔ the line is a secant of the circle. (ii) c2 = a2 (1 + m 2) ⇔ the line touches the circle. (It is tangent to the circle) (iii) c2 < a2 (1 + m 2) ⇔ the line does not meet the circle i. e. passes out side the circle. Example : For what value of c will the line y = 2x + c be a tangent to the circle x 2 + y2 = 5 ? Solution : We have : y = 2x + c or 2x – y + c = 0 ......(i) and x 2 + y2 = 5 ........(ii) If the line (i) touches the circle (ii), then length of the ⊥ from the centre (0, 0) = radius of circle (ii) ⇒ 2×0 − 0 + c 2 + ( −1) 2 Page : 3 of 30 , CIRCLE ∴ 2 g2 − c = 8 ⇒ 2 g2 − 9 = 8 ⇒ g2 – 9 = 16 ⇒ g= ±5 2 2 Hence, the required circle is x + y ± 10x + 6y + 9 = 0. Self Practice Problems : 1. Find the equation of a circle which touches the axis of y at a distance 3 from the origin and intercepts a distance 6 on the axis of x. 98930 58881 , BHOPAL, (M.P.) Let the equation of the circle be x 2 + y2 + 2gx + 2fy + c = 0. Since it touches y-axis at (0, –3) and (0, – 3) lies on the circle. ∴ c = f 2 ...(i) 9 – 6f + c = 0 .......(ii) (f – 3)2 = 0 ⇒ f = 3. From (i) and (ii), we get 9 – 6f + f 2 = 0 ⇒ Putting f = 3 in (i) we obtain c = 9. It is given that the circle x 2 + y2 + 2gx + 2fy + c = 0 intercepts length 8 on x-axis 2 = 5 ⇒ c 5 = 5 TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, Solution :
• 4. 6. Tang ent : Ans. 40, –10 (a) Slope form : y = mx + c is always a tangent to the circle x 2 + y2 = a2 if c2 = a2 (1 + m 2). Hence, equation  a 2m a 2  , . c c  of tangent is y = mx ± a 1 + m 2 and the point of contact is  − FREE Download Study Package from website: www.tekoclasses.com (b) Point form : (i) The equation of the tangent to the circle x 2 + y2 = a2 at its point (x 1, y1) is, x x 1 + y y1 = a². (ii) The equation of the tangent to the circle x 2 + y2 + 2gx + 2fy + c = 0 at its point (x 1, y1) is: xx 1 + yy1 + g (x+x 1) + f (y+y1) + c = 0. NOTE : In general the equation of tangent to any second degree curve at point (x 1, y1) on it can be obtained by x + x1 y + y1 x1y + xy1 replacing x 2 by x x 1, y2 by yy1, x by , y by , xy by and c remains as c. 2 2 2 (c) Parametric form : The equation of a tangent to circle x 2 + y2 = a2 at (a cos α, a sin α) is x cos α + y sin α = a.  a cos α + β a sin α + β   2 , 2  NOTE : The point of intersection of the tangents at the points P(α) & Q(β ) is  α −β   cos α − β cos 2  2   2 2 Example : Find the equation of the tangent to the circle x + y – 30x + 6y + 109 = 0 at (4, –1). Solution : Equation of tangent is  y + ( −1)  x+4  + 109 = 0  +6  4x + (–y) – 30  2  2    or 4x – y – 15x – 60 + 3y – 3 + 109 = 0 or –11x + 2y + 46 = 0 or 11x – 2y – 46 = 0 Hence, the required equation of the tangent is 11x – 2y – 46 = 0 Example : Find the equation of tangents to the circle x 2 + y2 – 6x + 4y – 12 = 0 which are parallel to the line 4x + 3y + 5 = 0 Solution : Given circle is x 2 + y2 – 6x + 4y – 12 = 0 .......(i) and given line is 4x + 3y + 5 = 0 .......(ii) Centre of circle (i) is (3, –2) and its radius is 5. Equation of any line 4x + 3y + k = 0 parallel to the line (ii) .......(iii) If line (iii) is tangent to circle, (i) then | 4.3 + 3( −2) + k | = 5 or |6 + k| = 25 42 + 3 2 or 6 + k = ± 25 ∴ k = 19, – 31 Hence equation of required tangents are 4x + 3y + 19 = 0 and 4x + 3y – 31 = 0 Self Practice Problem : 1. Find the equation of the tangents to the circle x 2 + y2 – 2x – 4y – 4 = 0 which are (i) parallel, (ii) perpendicular to the line 3x – 4y – 1 = 0 Ans. (i) 3x – 4y + 20 = 0 and 3x – 4y – 10 = 0 (ii) 4x + 3y + 5 = 0 and 4x + 3y – 25 = 0 7. Normal : If a line is normal / orthogonal to a circle then it must pass through the centre of the y +f circle. Using this fact normal to the circle x 2 + y2 + 2gx + 2fy + c = 0 at (x 1, y1) is; y − y1 = 1 (x − x 1). x1 + g Exercise : Find the equation of the normal to the circle x 2 + y2 – 5x + 2y – 48 = 0 at the point (5, 6). Solution : The equation of the tangent to the circle x 2 + y2 – 5x + 2y – 48 = 0 at (5, 6) is x+5 x+6  +2   – 48 = 0 ⇒ 10x + 12y – 5x – 25 + 2y + 12 – 96 = 0 5x + 6y – 5   2   2  ⇒ 5x + 14y – 109 = 0 14 5 ∴ Slope of the tangent = – ⇒ Slope of the normal = 5 14 Hence, the equation of the normal at (5, 6) is y – 6 = (14/5) (x – 5) ⇒ 14x – 5y – 40 = 0 Self Practice Problem : 1. Find the equation of the normal to the circle x 2 + y2 – 2x – 4y + 3 = 0 at the point (2, 3). Ans. x – y + 1 = 0 8. Pair of Tangents from a Point: The equation of a pair of tangents drawn from the point A (x 1, y1) to the circle x 2 + y2 + 2gx + 2fy + c = 0 is : SS1 = T². Where S ≡ x 2 + y2 + 2gx + 2fy + c ; S1 ≡ x 1² + y1² + 2gx 1 + 2fy1 + c T ≡ xx 1 + yy1 + g(x + x 1) + f(y + y1) + c. 98930 58881 , BHOPAL, (M.P.) =± 5 ⇒ c=±5 5 Hence, the line (i) touches the circle (ii) for c = ± 5 Self Practice Problem : 1. For what value of λ, does the line 3x + 4y = λ touch the circle x 2 + y2 = 10x. Page : 4 of 30 , CIRCLE c TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, ⇒
• 5. x= 3( y − 1) ± 9( y − 1)2 + 8(2y 2 − 4 y + 2) 4 FREE Download Study Package from website: www.tekoclasses.com or 4x – 3y + 3 = ± 25 y 2 − 50 y + 25 = ± 5(y – 1) ∴ Separate equations of tangents are x – 2y + 2 = 0 and 2x + y – 1 = 0 Self Practice Problems : 1. Find the equation of the tangents through (7, 1) to the circle x 2 + y2 = 25. Ans. 12x 2 – 12y2 + 7xy – 175x – 25y + 625 = 0 9. Length of a Tangent and Power of a Point: The length of a tangent from an external point (x 1, y1) to the circle 2 2 S ≡ x 2 + y2 + 2gx + 2fy + c = 0 is given by L = x1 + y1 + 2gx1 + 2f1 y + c = S1 . Square of length of the tangent from the point P is also called the power of point w.r.t. a circle. Power of a point w.r.t. a circle remains constant. Power of a point P is positive, negative or zero according as the point ‘P’ is outside, inside or on the circle respectively. Exercise : Find the length of the tangent drawn from the point (5, 1) to the circle x 2 + y2 + 6x – 4y – 3 = 0 Solution : Given circle is x 2 + y2 + 6x – 4y – 3 = 0 .........(i) Given point is (5, 1). Let P = (5, 1) Now length of the tangent from P(5, 1) to circle (i) = 5 2 + 12 + 6.5 − 4.1 − 3 = 7 Self Practice Problems : 1. Find the area of the quadrilateral formed by a pair of tangents from the point (4, 5) to the circle Ans. 8 sq. units x 2 + y2 – 4x – 2y – 11 = 0 and a pair of its radii. 2. If the length of the tangent from a point (f, g) to the circle x 2 + y2 = 4 be four times the length of the tangent from it to the circle x 2 + y2 = 4x, show that 15f 2 + 15g2 – 64f + 4 = 0 10. D irec to r Ci rc le: The locus of the point of intersection of two perpendicular tangents is called the director circle of the given circle. The director circle of a circle is the concentric circle having radius equal to original circle. Example : Find the equation of director circle of the circle (x – 2)2 + (y + 1)2 = 2. 2 times the Centre & radius of given circle are (2, –1) & 2 respectively. . Centre and radius of the director circle will be (2, –1) & 2 × 2 = 2 respectively. . ∴ equation of director circle is (x – 2)2 + (y + 1)2 = 4 ⇒ x 2 + y2 – 4x + 2y + 1 = 0 Ans. Self Practice Problems : 1. Find the equation of director circle of the circle whose diameters are 2x – 3y + 12 = 0 and x + 4y – 5 = 0 and area is 154 square units. Ans. (x + 3) 2 + (y + 2)2 = 98 Solution : 11. Cho rd of Conta ct: If two tangents PT 1 & PT 2 are drawn from the point P(x 1, y1) to the circle S ≡ x 2 + y2 + 2gx + 2fy + c = 0, then the equation of the chord of contact T 1T 2 is: xx 1 + yy1 + g (x + x 1) + f (y + y1) + c = 0. NOTE : Here R = radius; L = length of tangent. (a) Chord of contact exists only if the point ‘P’ is not inside. 2 LR (b) (c) (d) (e) Length of chord of contact T 1 T2 = R 2 + L2 . Area of the triangle formed by the pair of the tangents & its chord of contact =  2RL   Tangent of the angle between the pair of tangents from (x 1, y1) =  2  L − R2    Equation of the circle circumscribing the triangle PT1 T2 is: (x − x 1) (x + g) + (y − y1) (y + f) = 0. R L3 R 2 + L2 Page : 5 of 30 , CIRCLE 98930 58881 , BHOPAL, (M.P.) ∴ TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, Ex. : Find the equation of the pair of tangents drawn to the circle x 2 + y2 – 2x + 4y = 0 from the point (0, 1) Solution : Given circle is S = x 2 + y2 – 2x + 4y = 0 .......(i) Let P ≡ (0, 1) For point P, S1 = 02 + 12 – 2.0 + 4.1 = 5 Clearly P lies outside the circle and T ≡ x . 0 + y . 1 – (x + 0) + 2 (y + 1) i.e. T ≡ –x +3y + 2. Now equation of pair of tangents from P(0, 1) to circle (1) is SS1 = T2 or 5 (x 2 + y2 – 2x + 4y) = (– x + 3y + 2)2 or 5x 2 + 5y2 – 10x + 20y = x 2 + 9y2 + 4 – 6xy – 4x + 12y or 4x 2 – 4y2 – 6x + 8y + 6xy – 4 = 0 or 2x 2 – 2y2 + 3xy – 3x + 4y – 2 = 0 .......(ii) Note : Separate equation of pair of tangents : From (ii), 2x 2 + 3(y – 1) x – 2(2y2 – 4y + 2) = 0
• 6. FREE Download Study Package from website: www.tekoclasses.com Solution : xα + yβ – 12 = 0 ....... (iv) Now lines (iii) and (iv) are same, therefore, equations (iii) and (iv) are identical α β −12 18 ∴ = = ∴ α = 6, β = – 5 −3 − 10 5 18   Hence P =  6, −  5   Self Practice Problems : 1. Find the co-ordinates of the point of intersection of tangents at the points where the line 2x + y + 12 = 0 meets the circle x 2 + y2 – 4x + 3y – 1 = 0 Ans. (1, – 2) 2. Find the area of the triangle formed by the tangents drawn from the point (4, 6) to the circle x 2 + y2 = 25 405 √ 3 and their chord of contact. Ans. ; 4x + 6y – 25 = 0 52 12. Pole and Polar: (i) (ii) (iii) (iv) (v) Example : Solution : Example : Solution : If through a point P in the plane of the circle, there be drawn any straight line to meet the circle in Q and R, the locus of the point of intersection of the tangents at Q & R is called the Polar of the point P; also P is called the Pole of the Polar. The equation to the polar of a point P (x 1, y 1 ) w.r.t. the circle x 2 + y2 = a2 is given by xx 1 + yy 1 = a 2 , & if the circle is general then the equation of the polar becomes xx 1 + yy1 + g (x + x 1) + f (y + y1) + c = 0 i.e. T = 0. Note that if the point (x 1, y1) be on the circle then the tangent & polar will be represented by the same equation. Similarly if the point (x 1, y1) be outside the circle then the chord of contact & polar will be represented by the same equation.  Aa 2 Ba 2  . ,− Pole of a given line Ax + By + C = 0 w.r.t. circle x 2 + y2 = a2 is  −  C C    If the polar of a point P pass through a point Q , then the polar of Q passes through P. Two lines L1 & L2 are conjugate of each other if Pole of L1 lies on L2 & vice versa. Similarly two points P & Q are said to be conjugate of each other if the polar of P passes through Q & vice-versa. Find the equation of the polar of the point (2, –1) with respect to the circle x 2 + y 2 – 3x + 4y – 8 = 0 Given circle is x 2 + y 2 – 3x + 4y – 8 = 0 ............(i) Given point is (2, –1) let P = (2, –1). Now equation of the polar of point P with respect to circle (i) x +2  y − 1  +4   –8=0 x.2 + y(–1) – 3  2    2  or 4x – 2y – 3x – 6 + 4y – 4 – 16 = 0 or x + 2y – 26 = 0 Find the pole of the line 3x + 5y + 17 = 0 with respect to the circle x 2 + y2 + 4x + 6y + 9 = 0 Given circle is x 2 + y2 + 4x + 6y + 9 = 0 ............(i) and given line is 3x + 5y + 17 = 0 ............(ii) Let P(α, β) be the pole of line (ii) with respect to circle (i) Now equation of polar of point P(α, β) with respect to circle (i) is xα + yβ + 2(x + α) + 3(y + β) + 9 = 0 or (α + 2)x + (β + 3) y + 2α + 3β + 9 = 0 ............(iii) Now lines (ii) and (iii) are same, therefore, Page : 6 of 30 , CIRCLE Example : 98930 58881 , BHOPAL, (M.P.) Solution : Find the equation of the chord of contact of the tangents drawn from (1, 2) to the circle x 2 + y2 – 2x + 4y + 7 = 0 Given circle is x 2 + y2 – 2x + 4y + 7 = 0 .......(i) Let P = (1, 2) For point P (1, 2), x 2 + y2 – 2x + 4y + 7 = 1 + 4 – 2 + 8 + 7 = 18 > 0 Hence point P lies outside the circle For point P (1, 2), T = x . 1 + y . 2 – (x + 1) + 2(y + 2) + 7 i.e. T = 4y + 10 Now equation of the chord of contact of point P(1, 2) w.r.t. circle (i) will be 4y + 10 = 0 or 2y + 5 = 0 Tangents are drawn to the circle x 2 + y2 = 12 at the points where it is met by the circle x 2 + y2 – 5x + 3y – 2 = 0; find the point of intersection of these tangents. Given circles are S1 ≡ x 2 + y2 – 12 = 0 ....... (i) and S2 = x 2 + y2 – 5x + 3y – 2 = 0 ....... (ii) Now equation of common chord of circle (i) and (ii) is i.e. 5x – 3y – 10 = 0 ....... (iii) S1 – S 2 = 0 Let this line meet circle (i) [or (ii)] at A and B Let the tangents to circle (i) at A and B meet at P(α, β), then AB will be the chord of contact of the tangents to the circle (i) from P, therefore equation of AB will be TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, Example :
• 7. Equation of the Chord with a given Middle Point: FREE Download Study Package from website: www.tekoclasses.com The equation of the chord of the circle S ≡ x 2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M (x 1, y1) is xx 1 + yy 1 + g (x + x 1) + f (y + y1) + c = x 12 + y12 + 2gx 1 + 2fy1 + c which is designated by T = S1. NOTE : (i) The shortest chord of a circle passing through a point ‘M’ inside the circle, is one chord whose middle point is M. (ii) The chord passing through a point ' M ' inside the circle and which is at a maximum distance from the centre is a chord with middle point M. Ex. :Find the equation of the chord of the circle x 2 + y 2 + 6x + 8y – 11 = 0, whose middle point is (1, –1) Solution : Equation of given circle is S ≡ x 2 + y2 + 6x + 8y – 11 = 0 Let L ≡ (1, –1) For point L(1, –1), S1 = 12 + (–1)2 + 6.1 + 8(–1) – 11 = –11 and T ≡ x.1 + y (–1) + 3(x + 1) + 4(y – 1) – 11 i.e. T ≡ 4x + 3y – 12 Now equation of the chord of circle (i) whose middle point is L(1, –1) is T = S1 or 4x + 3y – 12 = –11 or 4x + 3y – 1 = 0 −4 + 1 3 Second Method : Let C be the centre of the given circle, thenC ≡ (–3, –4). L ≡ (1, –1) slope of CL= = − 3 −1 4 ∴ Equation of chord of circle whose middle point is L, is 4 (x – 1) [ chord is perpendicular to CL) ∴ y+1=– 3 or 4x + 3y – 1 = 0 Self Practice Problems : 1. Find the equation of that chord of the circle x 2 + y2 = 15, which is bisected at (3, 2) Ans.3x + 2y – 13 = 0 2. Find the co-ordinates of the middle point of the chord which the circle x 2 + y2 + 4x – 2y – 3 = 0 cuts off on the  3 1 − ,  line y = x + 2. Ans.  2 2 14. 15. Equation of the chord joining two points of circle : The equation of chord PQ to the circle x 2 + y2 = a2 joining two points P(α) and (β) on it is given by. The equation of a straight line joining two point α & β on the circle x 2 + y2 = a2 is α+β α +β α −β x cos + y sin = a cos . 2 2 2 Common Tangents to two Circles: Case (i) Number of Tangents Condition 4 common tangents (2 direct and 2 transverse) r1 + r2 < c 1 c2 . (ii) 3 common tangents. r1 + r2 = c 1 c2 . (iii) 2 common tangents. r1 − r2 < c 1 c2 < r1 + r2 (iv) 1 common tangent. r1 − r2 = c 1 c2. (v) No common tangent. c1 c 2 < r 1 − r2. (Here C1C2 is distance between centres of two circles.) Page : 7 of 30 , CIRCLE 98930 58881 , BHOPAL, (M.P.) 13. TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, α+2 β+3 2α + 3β + 9 = = 3 5 17 (i) (ii) (iii) From (i) and (ii), we get 5α + 10 = 3β + 9 or 5α – 3β = – 1 ............(iv) From (i) and (iii), we get 17α + 34 = 6α + 9β + 27 or 11α – 9β = –7 ............(v) Solving (iv) & (v), we get α = 1, β = 2 Hence required pole is (1, 2). Self Practice Problems : 1. Find the co-ordinates of the point of intersection of tangents at the points where the line 2x + y + 12 = 0 meets the circle x 2 + y2 – 4x + 3y – 1 = 0. Ans. (1, – 2) 2. Find the pole of the straight line 2x – y + 10 = 0 with respect to the circle x 2 + y2 – 7x + 5y – 1 = 0 3 3  ,  Ans. 2 2