If a circle C passing through the point (4, 0 | vedclass

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(A) Given equation of the circle: $x^2+y^2+4x-6y-12=0$.Let the center of this circle be $O_1 = (-2, 3)$ and radius $r_1 = \sqrt{(-2)^2 + 3^2 - (-12)}

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$\sqrt{5}$

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(A) Given equation of the circle: $x^2+y^2+4x-6y-12=0$.Let the center of this circle be $O_1 = (-2, 3)$ and radius $r_1 = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{4+9+12} = 5$.The center of the required circle $C$ lies on the line passing through $O_1(-2, 3)$ and the point of contact $P(1, -1)$.The slope of the line $O_1P$ is $m = \frac{-1-3}{1-(-2)} = \frac{-4}{3}$.The line passing through $P(1, -1)$ and the center of circle $C$ (let it be $(h, k)$) must be perpendicular to the common tangent at $P$. The slope of the normal line is $m' = \frac{-1}{-4/3} = \frac{3}{4}$.The equation of this normal line is $y+1 = \frac{3}{4}(x-1) \Rightarrow 3x-4y-7=0$.Since the center $(h, k)$ lies on this line,$3h-4k=7$.Also,the distance from $(h, k)$ to $P(1, -1)$ is the radius $R$,and the distance from $(h, k)$ to $(4, 0)$ is also $R$. Thus,$(h-1)^2 + (k+1)^2 = (h-4)^2 + k^2$.Expanding this: $h^2-2h+1 + k^2+2k+1 = h^2-8h+16 + k^2$.Simplifying: $6h+2k=14 \Rightarrow 3h+k=7$.Solving the system $3h-4k=7$ and $3h+k=7$ by subtracting: $5k=0 \Rightarrow k=0$.Then $3h=7 \Rightarrow h=7/3$.The radius $R = \sqrt{(7/3-1)^2 + (0+1)^2} = \sqrt{(4/3)^2 + 1^2} = \sqrt{16/9 + 1} = \sqrt{25/9} = 5/3$. Wait,re-evaluating the distance condition: the circle touches externally,so the distance between centers is $R+r_1$. The center of $C$ is at distance $R$ from $P(1, -1)$ along the normal. The normal line is $3x-4y-7=0$. The center $(h, k)$ is $(1+R(3/5), -1+R(-4/5))$ or similar. Using the distance to $(4, 0)$ as $R$: $(h-4)^2 + k^2 = R^2$. Substituting $h=1+3R/5, k=-1-4R/5$: $(1+3R/5-4)^2 + (-1-4R/5)^2 = R^2 \Rightarrow (3R/5-3)^2 + (4R/5+1)^2 = R^2 \Rightarrow 9R^2/25 - 18R/5 + 9 + 16R^2/25 + 8R/5 + 1 = R^2 \Rightarrow R^2 - 2R + 10 = R^2 \Rightarrow 2R = 10 \Rightarrow R = 5$.

If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1, -1)$,then the radius of $C$ is

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