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

Vedclass · Accepted Answer

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

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(A) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. Given $S_1: x^2+y^2-8x-6y+21=0$ and $S_2: x^2+y^2-2x-15=0$. 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 $y=2$: $(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$ $(1+4-8-12+21) + \lambda(5-2-15) = 0$ $6 + \lambda(-12) = 0 \Rightarrow \lambda = \frac{1}{2}$. Substituting $\lambda = \frac{1}{2}$ into the family equation: $(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$ Multiply by $2$: $2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$ $3x^2+3y^2-18x-12y+27 = 0$ Dividing by $3$: $x^2+y^2-6x-4y+9 = 0$.

Find the equation of a circle which passes through the point $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

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