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!

5834 views around the world

You can reuse this answer
Creative Commons License

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