Question

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines
$x^2-y^2-2x+4y-3=0,$ is

• A. $x^2+y^2-2x-4y+4=0$
• B. $x^2+y^2+2x+4y-4=0$
• C. $x^2+y^2-2x+4y+4=0$
• D. None of these

Answer 1

The correct option is A

$x^2+y^2-2x-4y+4=0$

Let the required equation of the circle be $(x-h{)}^2+(y-k{)}^2=a^2$

Comparing the given equation $x^2-2x+4y-3=0$
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

$=(\frac{hf-bg}{ab-h^2},\frac{gh-af}{ab-h^2})=(\frac{-1}{-1},\frac{-2}{-1})=(1,2)$

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

The equation of the required circle is

$(x-1{)}^2+(y-2{)}^2=a^2$

Since circle passes through (1, 1), we have
$1=a^2$

$\therefore$ Equation of the required circle :

$(x-1{)}^2+(y-2{)}^2=1$

$\Rightarrow x^2+y^2-2x-4y+4=0$