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Question

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x2 − y2 −2x + 4y − 3 = 0, is
(a) x2 + y2 − 2x − 4y + 4 = 0
(b) x2 + y2 + 2x + 4y − 4 = 0
(c) x2 + y2 − 2x + 4y + 4 = 0
(d) none of these

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Solution

(a) x2 + y2 − 2x − 4y + 4 = 0

Let the required equation of the circle be x-h2+y-k2=a2.

Comparing the given equation x2 − y2 −2x + 4y − 3 = 0 with ax2+by2+2hxy+2gx+2fy+c=0, we get:

a=1, b=-1, h=0, g=-1, f=2, c=-3

Intersection point = hf-bgab-h2, gh-afab-h2 = -1-1, -2-1=1, 2

Thus, the centre of the circle is 1, 2.

The equation of the required circle is x-12+y-22=a2.
Since circle passes through (1, 1), we have:
1=a2

∴ Equation of the required circle:
x-12+y-22=1
x2+y2-2x-4y+4=0

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