The equation of the circle passing through (1 | vedclass

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(C) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.Here

Vedclass · Accepted Answer

$x^2+y^2-6x-4y+9=0$

Vedclass · Answer

(C) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.Here,$S_1 = x^2+y^2-8x-6y+21$ and $S_2 = x^2+y^2-2x-15$.The equation is $(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$.Since the circle passes through $(1,2)$,substitute $x=1$ and $y=2$ into the equation:$(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$$(1+4-8-12+21) + \lambda(1+4-2-15) = 0$$6 + \lambda(-12) = 0$$6 - 12\lambda = 0$ $\Rightarrow 12\lambda = 6$ $\Rightarrow \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ back into the family equation:$(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$Multiply by $2$:$2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$$2x^2+2y^2-16x-12y+42 + x^2+y^2-2x-15 = 0$$3x^2+3y^2-18x-12y+27 = 0$Divide by $3$:$x^2+y^2-6x-4y+9 = 0$.

The equation of the circle passing through $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$ is:

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