Let x^{2}+y^{2}+Ax+By+C=0 be a circle passing | vedclass

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(A) The circle $x^{2}+y^{2}+Ax+By+C=0$ passes through $(0,6)$,so $0^{2}+6^{2}+A(0)+B(6)+C=0$,which gives $6B+C=-36$ (Equation $1$).The parabola is $y=

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$16$

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(A) The circle $x^{2}+y^{2}+Ax+By+C=0$ passes through $(0,6)$,so $0^{2}+6^{2}+A(0)+B(6)+C=0$,which gives $6B+C=-36$ (Equation $1$).The parabola is $y=x^{2}$. The slope of the tangent at $(2,4)$ is $\frac{dy}{dx} = 2x = 2(2) = 4$. The equation of the tangent is $y-4=4(x-2)$,which simplifies to $4x-y-4=0$ (Equation $2$).The equation of the tangent to the circle at $(2,4)$ is $x(2)+y(4)+A\frac{x+2}{2}+B\frac{y+4}{2}+C=0$,which simplifies to $(4+A)x+(8+B)y+(2A+4B+2C)=0$ (Equation $3$).Since the circle touches the parabola at $(2,4)$,the tangents are identical. Comparing coefficients of Equation $2$ and Equation $3$:$\frac{4+A}{4} = \frac{8+B}{-1} = \frac{2A+4B+2C}{-4} = k$.From $\frac{4+A}{4} = \frac{8+B}{-1}$,we get $-(4+A) = 32+4B$,so $A+4B=-36$ (Equation $4$).From $\frac{4+A}{4} = \frac{2A+4B+2C}{-4}$,we get $-(4+A) = 2A+4B+2C$,so $3A+4B+2C=-4$ (Equation $5$).Subtracting Equation $4$ from Equation $5$: $(3A+4B+2C) - (A+4B) = -4 - (-36)$ $\Rightarrow 2A+2C=32$ $\Rightarrow A+C=16$.

Let $x^{2}+y^{2}+Ax+By+C=0$ be a circle passing through $(0,6)$ and touching the parabola $y=x^{2}$ at $(2,4)$. Then $A+C$ is equal to

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