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# 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 ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$. Comparing the given equation x2 − y2 −2x + 4y − 3 = 0 with $a{x}^{2}+b{y}^{2}+2hxy+2gx+2fy+c=0$, we get: $a=1,b=-1,h=0,g=-1,f=2,c=-3$ Intersection point = $\left(\frac{hf-bg}{ab-{h}^{2}},\frac{gh-af}{ab-{h}^{2}}\right)$ = $\left(\frac{-1}{-1},\frac{-2}{-1}\right)=\left(1,2\right)$ Thus, the centre of the circle is $\left(1,2\right)$. The equation of the required circle is ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$. Since circle passes through (1, 1), we have: $1={a}^{2}$ ∴ Equation of the required circle: ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=1$ $⇒$ ${x}^{2}+{y}^{2}-2x-4y+4=0$

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