CircleMedium
Question
The equation of the circle passing through (1, 1) and the points of intersection of x2 + y2 + 13x - 3y = 0 and 2x2 + 2y2 + 4x - 7 y - 25 = 0 is
Options
A.4x2 + 4y2 + 30x - 10y = 25
B.4x2 + 4y2 + 30x - 13y - 25 = 0
C.4x2 + 4y2 - 17x - 10y + 25 = 0
D.None of the above
Solution
The required equation of circle is,
(x2 + y2 + 13x - 3y) + λ
= 0 ....(i)
Its passing throught (1, 1)
⇒ 12 + λ(24) = 0
⇒ λ =
On putting in Eq. (i), we get
x2 + y2 + 13x - 3y
⇒ 4x2 + 4y2 + 52x - 12y - 22x - y - 25 = 0
⇒ 4x2 + 4y2 + 30x - 13y - 25 = 0
(x2 + y2 + 13x - 3y) + λ
= 0 ....(i)Its passing throught (1, 1)
⇒ 12 + λ(24) = 0
⇒ λ =
On putting in Eq. (i), we get
x2 + y2 + 13x - 3y
⇒ 4x2 + 4y2 + 52x - 12y - 22x - y - 25 = 0
⇒ 4x2 + 4y2 + 30x - 13y - 25 = 0
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