Question

Question #c7e0f

Answer 1

The equation of the Circle is

`(x - 2)^2 + (y - 3)^2 = 10`

Explanation:

Let the center of the circle be at a point (a,b)
As you may know the equation of a circle whose center is at (a,b) is
`(x - a)^2 + (y - b)^2 = r^2`
where r is the radius of the circle
It is given that the circle passes through the point (3,6)
Substituting in the above equation,
`(3 - a)^2 + (6 - b)^2 = r^2` . . . . . . (1)
It is also given that the circle passes through the point (5,4)
Substituting in the above equation,
`(5 - a)^2 + (4 - b)^2 = r^2` . . . . . . . (2)
(1) - (2) gives
`{(a^2 - 6 * a + 9 + b^2 - 12 * b + 36) - (a^2 - 10 * a + 25 + b^2 - 8 * b + 16)} = 0`
Cancelling `a^2 and b^2` and simplifying, we get
`(- 6 * a + 9 - 12 * b + 36 + 10 * a - 25 + 8 * b - 16) = 0`
or `4a - 4b + 4 = 0`
ie `b - a = 1` . . . . .(3)
The center lies on the line
`x + y - 5 = 0`
Substituting a for x and b for y, we get
`a + b - 5 = 0`
or `a + b = 5` . . . .(4)
(3) + (4) gives,
2b = 6 or b = 3
Substituting for b = 3 in (4), we get a = 5 - 3 = 2
So the center of the circle is at (2,3)
Substituting these values in (1), we get
`(3 - 2)^2 + (6 - 3)^2 = 10 = r^2`

Thus the equation of the circle is
`(x - 2)^2 + (y - 3)^2 = 10`