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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

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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 $$ \left(x+\frac32\right)^2+(y-4)^2=\frac{25}4. $$

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Saying that "Another circle C2 touches the first circle at a point P on the positive y-axis" means "at the point where C1 crosses the positive y-axis" which is easily found to be (0, 6). It "passes through the center of C1" means that (-3, 2) is another point on the circle. That means that the perpendicular bisector of the line segment between (0, 6) and (-3, 2) passes through the center of C2. The midpoint is (-3/2, 4) and the slope of that line is (2- 6)/(-3- 0)= 4/3. The slope of a perpendicular is -3/4 and the line with slope -3/4, passing thorough (-3/2, 4), is y= -3/4(x- 3/2)+ 4= -(3/4)x+ 7/8 or 3x+ 8y= 7. The radius of C1 is 5 so the radius of C2 is 5/2. The center of C2 is the point where 3x+ 8y= 7 and $(x- 3/2)^2+ (y- 4)^2= 25/4$.

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