A circle C has equation $x^2+y^2-6x+8y-75=0$ , and a second circle has a centre at $(15,12)$ and radius 10. What are the coordinates of the point where they touch?

Question

I worked out C as being $(x-3)^2+(y+4)^2=100$ and the second circle as $(x-15)^2+(y-12)^2=100$ . I graphed the answer and got $(9,4)$ as the answer but I'm not sure how to get there. Thanks!

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Answer 1 · 2017-11-30T20:49:08

Set the equations equal to each other

You did a great job getting to $(x-3)^2+(y+4)^2 =100 and (x-15)^2+(y+12)^2 = 100$

Both side equal 100 so set $(x-3)^2+(y+4)^2 = (x-15)^2+(y+12)^2$

$x^2-6x+9+y^2+8x+16=x^2-30x+225+y^2-24y+144$

Simplify to 24x = 216, x = 9

32y = 128, y = 4