# Question in analytical geometry

1. Dec 30, 2003

### Chen

The point (1, 0) is on a circle, which center is on the line y = -2x. The line 3x + 4y + 15 = 0 is a tangent to the circle. Find the equation(s) for the circle(s).

Thanks,

2. Dec 30, 2003

### Muzza

The equation of a circle is (x - a)^2 + (y - b)^2 = r^2.

The centre is on the line y = -2x, which means b = -2a, hence our circle has the equation:

(x - a)^2 + (y - (-2a))^2 = r^2
<=>
(x - a)^2 + (y + 2a)^2 = r^2

Since (1, 0) was on the circle:

(1 - a)^2 + (0 + 2a)^2 = r^2
<=>
r^2 = 5a^2 - 2a + 1

Putting that back into the equation of our circle:

(x - a)^2 + (y + 2a)^2 = 5a^2 - 2a + 1 ... (1)

We know that (1) must intersect with 3x + 4y + 15 = 0 somewhere, find that point. We know that at that point, the derivative of our circle's equation with respect to x must be -3/4 (i.e the slope of the tangent line), which gives us an equation where we can solve for a.

But I'll be damned if I'm going to do all those nasty calculations by hand ;) I worked them out with my computer, which gave a = -2 and a = 1, so the equations of the circles might be these:

(x - 1)^2 + (y + 2)^2 = 4
(x + 2)^2 + (y - 4)^2 = 25

They seem to work...

Last edited: Dec 30, 2003
3. Dec 30, 2003

### Chen

Thanks for solving this. We are actually not allowed to use things like derivative calculations in this type of questions, so I took a different approach. But the answers are the same so I'm happy. :)