Q.359. Equation of a circle given two points and a line containing center

Question 359.
Find the equation of the circle which passes through two points (2,0) and (0,4) and has centre lying on the line x+2y+4=0.

Answer 359.
Let the center be (h, k).
The center lies on the line x + 2y + 4 = 0
=> h + 2k + 4 = 0 ... ( 1 )
 Also, the center must be equidistant from the points
on the circle (2, 0) and (0, 4)
=> (h - 2)^2 + k^2 = h^2 + (k - 4)^2
=> 4h - 8k + 12 = 0
=> h - 2k + 3 = 0 ... ( 2 )

Adding ( 1 ) and ( 2 ),
2h = - 7
=> h = - 7/2

Plugging in eqn. ( 1 ),
- 7/2 + 2k + 4 = 0
=> k = (1/2) (7/2 - 4) = - 1/4
=> center of the circle = (- 7/2, - 1/4)
Distance of the center from a point (2, 0) on the circle = radius r
=> r^2 = (2 + 7/2)^2 + (0 + 1/4)^2 = 485/16

The equation of the circle having center (- 7/2, - 1/4)
and r^2 = 485/16 is
(x + 7/2)^2 + (y + 1/4)^2 = 485/16
=> x^2 + y^2 + 7x + (1/2)y + 49/4 + 1/16 = 485/16
=> 2x^2 + 2y^2 + 14x + y + 49/2 + 1/8 - 485/8 = 0
=> 2x^2 + 2y^2 + 14x + y - 36 = 0.

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