Two circles, one centered at $(-3,2)$ and the other centered at $(0,-1)$, are internally tangent as shown.
Link to diagram: https://latex.artofproblemsolving.com/7/d/8/7d8d7be23ae4f2eecfa363a19ed7b05d06f0c7be.png
If the equation of the smaller circle can be written as $x^2 + y^2 + Dx + Ey + F = 0$, find $D + E + F$.
radius of big circle = distance between (0, -1) and (1, 6) .
radius of big circle = \(\sqrt{(1-0)^2+(6--1)^2}=\sqrt{50}=5\sqrt{2}\)
Draw a line from (0, -1) to (-3, 2) . This is the blue line.
length of blue line = \(\sqrt{(-3 - 0)^2+(2--1)^2}=\sqrt{18}=3\sqrt2\)
Then, draw a continuation of that line from (-3, 2) to the point on both circles. This is the red line.
length of red line = \(5\sqrt2-3\sqrt2=2\sqrt2\)
We know the center of the small circle, and we know the length of the red line, which is a radius of the small circle. So, we can write the equation of the small circle like this...
(x - -3)2 + (y - 2)2 = (2√2)2 Multiply out the parenthesees.
(x + 3)(x + 3) + (y - 2)(y - 2) = (2√2)(2√2)
x2 + 6x + 9 + y2 - 4y + 4 = 8
x2 + 6x + y2 - 4y + 13 = 8
x2 + y2 + 6x - 4y + 5 = 0
And...... 6 - 4 + 5 = 7
radius of big circle = distance between (0, -1) and (1, 6) .
radius of big circle = \(\sqrt{(1-0)^2+(6--1)^2}=\sqrt{50}=5\sqrt{2}\)
Draw a line from (0, -1) to (-3, 2) . This is the blue line.
length of blue line = \(\sqrt{(-3 - 0)^2+(2--1)^2}=\sqrt{18}=3\sqrt2\)
Then, draw a continuation of that line from (-3, 2) to the point on both circles. This is the red line.
length of red line = \(5\sqrt2-3\sqrt2=2\sqrt2\)
We know the center of the small circle, and we know the length of the red line, which is a radius of the small circle. So, we can write the equation of the small circle like this...
(x - -3)2 + (y - 2)2 = (2√2)2 Multiply out the parenthesees.
(x + 3)(x + 3) + (y - 2)(y - 2) = (2√2)(2√2)
x2 + 6x + 9 + y2 - 4y + 4 = 8
x2 + 6x + y2 - 4y + 13 = 8
x2 + y2 + 6x - 4y + 5 = 0
And...... 6 - 4 + 5 = 7