If the circle passing through (3,5), (5,5) an | vedclass

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(D) Let the equation of the circle be $x^2+y^2+2gx+2hy+c=0$. Since it passes through $(3,5), (5,5)$, and $(3,-3)$, we have: $9+25+6g+10h+c=0 \implies

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-$4$

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(D) Let the equation of the circle be $x^2+y^2+2gx+2hy+c=0$. Since it passes through $(3,5), (5,5)$, and $(3,-3)$, we have: $9+25+6g+10h+c=0 \implies 6g+10h+c=-34$ $25+25+10g+10h+c=0 \implies 10g+10h+c=-50$ $9+9+6g-6h+c=0 \implies 6g-6h+c=-18$ Subtracting the first from the second: $4g=-16 \implies g=-4$. Subtracting the third from the first: $16h=-16 \implies h=-1$. Substituting $g$ and $h$ into the first equation: $6(-4)+10(-1)+c=-34 \implies -24-10+c=-34 \implies c=0$. The circle is $x^2+y^2-8x-2y=0$. Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ are orthogonal if $2g_1g_2+2f_1f_2=c_1+c_2$. Here, $g_1=-4, f_1=-1, c_1=0$ and $g_2=1, f_2=f, c_2=0$. $2(-4)(1)+2(-1)(f)=0+0 \implies -8-2f=0 \implies 2f=-8 \implies f=-4$.

If the circle passing through $(3,5), (5,5)$ and $(3,-3)$ cuts the circle $x^2+y^2+2x+2fy=0$ orthogonally, then $f=$

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