The equation of the circle whose diameter is the common chord of the circles x^2+y^2+2x+2y+1=0 and x^2+y^2+4x+6y+4=0 is

Question

Accepted Answer

Correct answer: A. (a) Given, S_1 ≡x^2+y^2+2x+2y+1=0 and S_2 ≡x^2+y^2+4x+6y+4=0 Now, the equation of common chord of these two circles. S_1-S_2=0(x^2+y^2+2x+2y+1)-(x^2+y^2+4x+6y+4)=0 ⇒-2x-4y-3=0 i .e . 2x+4y+3=0 The equation of required circle, S_1+ (S_2-S_1)=0(x^2+y^2+2x+2y+1)+ (2x+4y+3 )=0 x^2+y^2+2x (1+ )+2y (2 +1 )+3 +1=0 Diameter of the above circle is the common chord of circles S_1 and S_2. So , 2 (- (1+ ) )+4 (- (2 +1 ) )+3=0 ⇒-2-2 -8 -4+3=0 ⇒10 =-3 ⇒ = -310 Hence, the required circle x^2+y^2+2x(1- 310)+2y(- 610+1)- 910+1=0x^2+y^2+ 14x10+ 8y10+ 110=010x^2+10y^2+14x+8y+1=0

$The equation of the circle whose diameter is the common chord of the circles x^{2} + y^{2} + 2 x + 2 y + 1 = 0 and x^{2} + y^{2} + 4 x + 6 y + 4 = 0 is$

Solution

More Hyperbola PYQs