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(C) The center $(h, k)$ of the circle is the intersection of the diameters $x+2y-2=0$ and $2x+3y-1=0$. Solving these equations: $x+2y=2$ $(1)$ $2x+3y=

Vedclass · Accepted Answer

$-44$

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(C) The center $(h, k)$ of the circle is the intersection of the diameters $x+2y-2=0$ and $2x+3y-1=0$. Solving these equations: $x+2y=2$ $(1)$ $2x+3y=1$ $(2)$ Multiplying $(1)$ by $2$: $2x+4y=4$ $(3)$ Subtracting $(2)$ from $(3)$: $(2x+4y)-(2x+3y) = 4-1$,so $y=3$. Substituting $y=3$ in $(1)$: $x+2(3)=2 \implies x+6=2 \implies x=-4$. So,the center is $(-4, 3)$. The radius $R$ is the distance between the center $(-4, 3)$ and the point $(8, 8)$: $R^2 = (8 - (-4))^2 + (8 - 3)^2 = 12^2 + 5^2 = 144 + 25 = 169$. The equation of the circle is $(x - h)^2 + (y - k)^2 = R^2$: $(x + 4)^2 + (y - 3)^2 = 169$ $x^2 + 8x + 16 + y^2 - 6y + 9 = 169$ $x^2 + y^2 + 8x - 6y - 144 = 0$. Comparing with $x^2 + y^2 + px + qy + r = 0$,we get $p=8, q=-6, r=-144$. Then $p^2 + q^2 + r = 8^2 + (-6)^2 - 144 = 64 + 36 - 144 = 100 - 144 = -44$.

If the equation of the circle passing through the point $(8,8)$ and having the lines $x+2y-2=0$ and $2x+3y-1=0$ as its diameters is $x^2+y^2+px+qy+r=0$,then $p^2+q^2+r=$

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