Example 1 :
Find the equation of the circle with center (2, 3) and passing through the intersection of the lines 3x − 2y −1 = 0 and 4x + y − 27 = 0
Solution :
By finding the point of intersection of the lines, we get a point lies on the circle.
3x − 2y − 1 = 0 ----------(1)
4x + y − 27 = 0 ----------(2)
3x − 2y − 1 = 0
(2) x 2 ==> 8x + 2y − 54 = 0
--------------------
11x - 55 = 0 ==> x = 5
Apply x = 5 in (1) to get the value of y.
3(5) - 2y - 1 = 0
15 - 2y - 1 = 0
-2y = -14
y = 7
So, the point lies on the circle is (5, 7).
By finding the distance between center and a point lies on the circle, we get the radius.
r = √(2 - 5)2 + (3 - 7)2
r = √(-3)2 + (-4)2 = √25
r = 5
Equation of the circle :
(x - h)2 + (y - k)2 = r2
(h, k) ==> (2, 3)
(x - 2)2 + (y - 3)2 = 52
x2 - 4x + 4 + y2 - 6y + 9 - 25 = 0
x2 + y2 - 4x - 6y - 12 = 0
Example 2 :
Obtain the equation of the circle for which (3, 4) and (2,-7) are the ends of a diameter.
Solution :
Equation of circle with end points of diameter :
(x − x1)(x − x2) + ( y − y1)( y − y2) = 0
(x1, y1) ==> (3, 4) and (x2, y2) ==> (2, -7)
(x − 3)(x − 2) + ( y − 4)( y + 7) = 0
x2 - 5x + 6 + y2 + 3y - 28 = 0
x2 + y2 - 5x + 3y + 6 -28 = 0
x2 + y2 - 5x + 3y - 22 = 0
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