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

Answer 1

(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:
${\left(x-1\right)}^2+{\left(y-2\right)}^2=1$
$\Rightarrow$ $x^2+y^2-2x-4y+4=0$