SOLUTION: A circle passes through the centre of a circle X^2 + Y^2 - 3X - 4Y - 1 = 0 and it's centre is (5,2).find it's equation.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A circle passes through the centre of a circle X^2 + Y^2 - 3X - 4Y - 1 = 0 and it's centre is (5,2).find it's equation.       Log On


   

Question 885215: A circle passes through the centre of a circle X^2 + Y^2 - 3X - 4Y - 1 = 0 and it's centre is (5,2).find it's equation.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
For the first circle, the given equation is x%5E2%2By%5E2-3x-4y-1=0 .
We can transform it:
x%5E2%2By%5E2-3x-4y=1
x%5E2-3x%2By%5E2-4y=1
x%5E2-3x%2B9%2F4%2By%5E2-4y%2B4=1%2B9%2F4%2B4
%28x-3%2F2%29%5E2%2B%28y-2%29%5E2=1%2B9%2F4%2B4

x=3%2F2 and y=2 are the coordinates of the center of the circle,
and %28x-3%2F2%29%5E2%2B%28y-2%29%5E2 is the square of the distance from a point (x,y) to that center.
1%2B9%2F4%2B4 is the square of the radius of the circle, but I do not need to calculate it.

The distance from the center of that first circle (with x=3%2F2 and y=2 )
to point (5,2) (also with y=2 ) is abs%285-3%2F2%29=abs%287%2F2%29=7%2F2 .

A circle centered at (5,2), with radius R has the equation
%28x-5%29%5E2%2B%28y-2%29%5E2=R%5E2
If the circle passes through a point with x=3%2F2 and y=2 ,
which is at a distance 7%2F2 from (5,2), then R=7%2F2 , and
%28x-5%29%5E2%2B%28y-2%29%5E2=%287%2F2%29%5E2 is the equation of the second circle.
It can also be written as
%28x-5%29%5E2%2B%28y-2%29%5E2=49%2F4 , or further manipulated into whatever equivalent equation you desire:
x%5E2-10x%2B25%2By%5E2-4y%2B4=49%2F4-->x%5E2%2By%5E2-10x-4y%2B25%2B4-49%2F4=0-->x%5E2%2By%5E2-10x-4y%2B67%2F4=0-->4x%5E2%2B4y%5E2-40x-16y%2B67=0