Question

The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y=T$ is

• A. $3x^2+3y^2-8x-13y=0$
• B. $3x^2+3y^2-8x+29y=0$
• C. $3x^2+3y^2+8x+29y=0$
• D. $3x^2+3y^2-8x-29y=0$

Answer 1

Answer: (B)

$3x^2+3y^2-8x+29y=0$

Let $c:x^2+y^{2}0+2gx+2fy=0$ be the equation of circle which passes through origin

$C=(-g,-f),r=\sqrt{g^2+f^2}$

$C1:x^2+y^{2}0-6x+8=0$

$C1=(3,0),r1=\sqrt{9-8}=1$

$C2:x^2+y^{2}0-2x–2y=7$

$C1=(1,2),r1=\sqrt{1+1+7}=3$

Given c is orthogonal to c1 and c2

$d^2=r{1}^2+r{2}^2$

$⟹(s+g{)}^2+d^2=(g+f{)}^2+1^2$

$⟹g+6g=1$

$⟹6g=-8⟹g=\frac{-4}{3}$

$(1+g{)}^2+(1+f{)}^2=(g^2+f^2)$

$1+2g+1+2f=g$

$⟹g+f=\frac{7}{2}$

$⟹f=\frac{7}{2}+\frac{4}{3}=\frac{29}{6}$

Equation of circle is

$x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$

$3x^2+3y^2–8x+29y=0$