Equation of circle given center and tangent line

Question

What is the equation for a circle which has its center at (1, 2) and is tangent to the line -3x+2y= 14. Solve using your algebra skills; DO NOT USE CONSTRUCTIONS!!!!

 

The equation of a circle is r²=(x-h)²+(y-k)². So far I have r²=(x-1)²+(y-2)². I can rearrange the tangent line equation and get (3/2)x+7=y. I just need to find r. I am wondering if I can substitute (3/2)x+7 for y in the equation of a circle to get r. If I cannot, what should I do instead? If I can, is there a better way to do this?

Answer 1

Yes there is a better way.   The hint here is that the circle and line intersect at only one point.   So your goal is to find a quadratic equation that results in a single root.

 

(x-1)² + ((3/2)x +7 - 2)² - r² = 0  

 

x² - 2x + 1 + 2.25X² + 15x + 25 -r² = 0

 

3.25X2 + 13X + (26-r²) = 0

 

Factor out the 3.25  

 

3.25(X2 + 4x + (26-r²)/3.25)   = 0;

 

In order to get a single root, the quadratic expression inside of the parenthesis must be a perfect square. 

 

b² - 4ac = 0;

 

 

16 - 4 * 1 *   (26-r²)/3.25 = 0

16 = 4/3.25(26-r²)

 

16 = 32 - 4r²/3.25

 

-16 = -4r²/3.25

 

4*3.25 = r^{2 = 13}

 

(x -1)² + (y-2)² = 13