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?

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!

Answer 1

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

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Answer 2

To find the point where the two circles touch, we need to solve for the intersection of the two circles. First, complete the square for the equation of circle C to rewrite it in standard form. Then, use the distance formula to find the distance between the centers of the two circles. If this distance equals the sum of the radii of the circles, they touch at one point. Use this distance to find the coordinates of the point of intersection.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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