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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 $D + E + F$.

michaelcai  Jul 24, 2017

Best Answer 

 #1
avatar+4711 
0

 

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

hectictar  Jul 24, 2017
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2+0 Answers

 #1
avatar+4711 
0
Best Answer

 

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

hectictar  Jul 24, 2017
 #2
avatar+76901 
+1

 

Nice, hectictar....!!!!

 

 

 

cool cool cool

CPhill  Jul 25, 2017

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