A circle C1 has the equation $(x+3)^2 + (y-2)^2 = 25$. Another circle C2 touches the first circle at a point P on the positive y-axis and passes through the centre of C1. The diameter of C1 is twice the diameter of C2. Find the equation of C2
[Math] Two circles touch internally. Find equation of smaller circle given equation of large circle
circles
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Best Answer
The point $P$ is given by $x=0$ and $3^2+(y-2)^2=25$, so $(y-2)^2=16$, or $y=2\pm 4$. As the positive choice is desired, $y=6$. The radius of $C_2$ is $5/2$, and it passes through $(0,6)$ and $(-3,2)$.
From the equation $(x-x_0)^2+(y-y_0)^2=25/4$ we get $$ x_0^2+(6-y_0)^2=\frac{25}4,\ \ \ \ (x_0+3)^2+(2-y_0)^2=\frac{25}4. $$ Equating the two equations the squares cancel, so we get $$ 36-12y_0=6x_0+9-4y_0+4, $$ or $$ 6x_0+8y_0=23. $$ Now one can substitute and solve.
A more heuristic way to do the problem is to draw the picture and to notice that putting the centre in the middle point between $(-3,2)$ and $(0,6)$ fulfills the specs. So we can take $$ x_0=-\frac32,\ \ y_0=\frac{2+6}2=4, $$ and then $C_2$ would be given by $$ \left(x+\frac32\right)^2+(y-4)^2=\frac{25}4. $$