Equations of Circles

Question

Find the equation of the circle which passes through (-2,1) and is tangent to the line 3x-2y=6 at the point (4,3).

Answer 1

The general formula for a circle is (x- h)² + (y - k)²= r²

Now we have to solve for h, k, and r

It goes through the point (-2, 1), so (-2- h)² + (1-k)² = r²

It also goes through the point (4,3), so (4- h)² + (3-k)² = r²

Lastly, it's tangent to the line 3x-2y=6 at (4,3). This line has a slope of -3/2

Take the derivative of the circle with respect to x.

2(x-h) + 2(y-k)(dy/dx) = 0

dy/dx = -3/2 at (4,3), so plug those values in and simplify.

2(4-h) + 2(3-k)(-3/2) = 0

8 - 2h - 9 + 3k = 0

3k - 2h = 1

We have 3 equations now with 3 unknowns so this is a system of equations:

(-2- h)² + (1-k)² = r²

(4- h)² + (3-k)² = r²

3k - 2h = 1

Do you think you can solve for h, k, and r