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(C) The equation of a circle passing through the intersection of circles $S_1: x^2+y^2-8x=0$ and $S_2: x^2+y^2-9=0$ is given by $S_1 + \lambda S_2 = 0

Vedclass · Accepted Answer

$32x^2+32y^2-72x-207=0$

Vedclass · Answer

(C) The equation of a circle passing through the intersection of circles $S_1: x^2+y^2-8x=0$ and $S_2: x^2+y^2-9=0$ is given by $S_1 + \lambda S_2 = 0$,where $\lambda 
eq -1$. $(x^2+y^2-8x) + \lambda(x^2+y^2-9) = 0$ $(1+\lambda)x^2 + (1+\lambda)y^2 - 8x - 9\lambda = 0$ $x^2 + y^2 - \frac{8}{1+\lambda}x - \frac{9\lambda}{1+\lambda} = 0$. The center of this circle is $C\left(\frac{4}{1+\lambda}, 0ight)$. The equation of the common chord is $S_1 - S_2 = 0$,which is $(x^2+y^2-8x) - (x^2+y^2-9) = 0$,i.e.,$8x = 9$. Since the common chord is the diameter,the center $C$ must lie on the line $8x = 9$. $8\left(\frac{4}{1+\lambda}ight) = 9$ $32 = 9(1+\lambda) = 9 + 9\lambda$ $9\lambda = 23 \Rightarrow \lambda = \frac{23}{9}$. Substituting $\lambda = \frac{23}{9}$ into the equation $(1+\lambda)x^2 + (1+\lambda)y^2 - 8x - 9\lambda = 0$: $(1 + \frac{23}{9})x^2 + (1 + \frac{23}{9})y^2 - 8x - 9(\frac{23}{9}) = 0$ $\frac{32}{9}x^2 + \frac{32}{9}y^2 - 8x - 23 = 0$ Multiplying by $9$,we get $32x^2 + 32y^2 - 72x - 207 = 0$. Thus,option $C$ is correct.

The equation of the circle having the common chord of the circles $x^2+y^2-8x=0$ and $x^2+y^2-9=0$ as its diameter is

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