find the equation of the circles described. tangent to 2x + 3y + 13 = 0 and 2x - 3y - 1 = 0 and contains (0,4)

Answer 1

Let L1 be the line 2x + 3y + 13 = 0. Its slope is -2/3
Let L2 be the line 2x - 3y - 1 = 0. It has slope 2/3.

So, the bisectors of the angles formed by L1,L2 are horizontal and vertical. Since (0,4) is above both lines, we want the vertical bisector x = -3.

So, we want a circle whose center is (-3,k) which passes through (0,4) and is tangent to L1 (and hence L2).

The distance from (-3,k) to L1 is (3k+7)/√13

So far we have the circle

(x+3)^2 + (y-k)^2 = (3k+7)^2/13

We know the circle passes through (0,4), so

3^2 + (4-k)^2 = (3k+7)^2/13
k=2

Looks like the circle is
(x+3)^2 + (y-2)^2 = 13