We are given A(3,4), B(7,-6). A circle has diameter AB. a) Find an equation for this circle b)Another circle have the same radius and is tangent to point A. What is the circle equation?

I found out that
a) (x-5)^2+(y+1)^2=29

But I am stuck at b

1 Answer
Apr 24, 2018

a) (x-5)^2+(y+1)^2=29
b) (x-1)^2+(y-9)=29

Explanation:

a) We first know that the equation (x-a)^2+(y-b)^2=r^2 is the equation for a circle where a is the x coordinate of the center of the circle, b is the y coordinate of the center of the circle, and r is the radius of the circle.

We can find the center coordinate of the circle by adding the x and y coordinates of A and B against each other and then dividing by 2. For the x coordinate, we do (3+7)/2 to get 5 as the x coordinate for the center. We now do the same for the y coordinate to get (4-6)/2=-1.

We now replace a and b to now get the equation (x-5)^2+(y+1)^2=r^2. Now all we have to do is find r. We use the Pythagorean Theorem to find that the length of AB is sqrt((7-3)^2+(4+6)^2)=sqrt(116)=2sqrt(29). If we know that the diameter is 2sqrt(29), then the radius is sqrt(29). We replace r with 29 to get (x-5)^2+(y+1)^2=29.

b) We name the center of the new circle C. The center of the old circle should be named D. We know that A must be the center of CD. We find that the center of the new circle is ( 1 , 9 ) and because the radius is the same, the new equation is (x-5)^2+(y+1)^2=29.