The normal to a circle S=0 at P(1,3) is x+2y= | vedclass

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(A) The polar of the point $A(7, -1/2)$ with respect to the circle $x^2+y^2-4x+6y-12=0$ is given by $T=0$: $7x - (1/2)y - 2(x+7) + 3(y - 1/2) - 12 = 0

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$x^2+y^2-10x-2y+6=0$

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(A) The polar of the point $A(7, -1/2)$ with respect to the circle $x^2+y^2-4x+6y-12=0$ is given by $T=0$: $7x - (1/2)y - 2(x+7) + 3(y - 1/2) - 12 = 0$ $7x - 0.5y - 2x - 14 + 3y - 1.5 - 12 = 0$ $5x + 2.5y - 27.5 = 0$ Multiplying by $2/5$,we get $2x + y - 11 = 0$,or $2x + y = 11$. Since the normals to a circle always pass through its center,the center $(h, k)$ is the intersection of the two normals: $x + 2y = 7$ $(i)$ $2x + y = 11$ (ii) Multiplying $(i)$ by $2$: $2x + 4y = 14$. Subtracting (ii) from this: $3y = 3 \Rightarrow y = 1$. Substituting $y=1$ in $(i)$: $x + 2(1) = 7 \Rightarrow x = 5$. So,the center is $(5, 1)$. The radius $r$ is the distance from $(5, 1)$ to $P(1, 3)$: $r^2 = (5-1)^2 + (1-3)^2 = 4^2 + (-2)^2 = 16 + 4 = 20$. The equation of the circle is $(x-5)^2 + (y-1)^2 = 20$. $x^2 - 10x + 25 + y^2 - 2y + 1 = 20$ $x^2 + y^2 - 10x - 2y + 6 = 0$.

The normal to a circle $S=0$ at $P(1,3)$ is $x+2y=7$ and it has another normal at $Q(3,5)$ which is the polar of the point $A(7, -1/2)$ with respect to the circle $x^2+y^2-4x+6y-12=0$. Then,the equation of the circle $S=0$ is

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