Question

Let $C$ be a circle passing through the origin and making an intercept of $\sqrt{10}$ on the line $y=2x+\frac{5}{\sqrt{2}}$. If the line subtends an angle of ${45}^{\circ }$ at the origin, then the equation of circle $C$ is/are

• A. $x^2+y^2-4x-2y=0$
• B. $x^2+y^2-2x-4y=0$
• C. $x^2+y^2+4x+2y=0$
• D. $x^2+y^2+2x+4y=0$

Answer 1

Answer: (B), (D)

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

Let $C:x^2+y^2+2gx+2fy=0$ and $r$ be the radius of the circle, so
$g^2+f^2=r^2\cdots \left(1\right)$
Given line is $y=2x+\frac{5}{\sqrt{2}}$
As the line subtends ${45}^{\circ }$ at the circumference(origin) of the circle, so
$\angle PCQ={90}^{\circ }$


$\angle CPQ={45}^{\circ }\Rightarrow \sin \angle CPQ=\frac{r}{\sqrt{10}}\Rightarrow r=\sqrt{5}$
Using equation $\left(1\right)$, we get
$g^2+f^2=5\cdots \left(2\right)$
$\sin \angle CPQ=\frac{CL}{\sqrt{5}}\Rightarrow CL=\frac{\sqrt{5}}{\sqrt{2}}$
Perpendicular from centre, we get
$\Rightarrow \frac{|f-2g+\frac{5}{\sqrt{2}}|}{\sqrt{1+4}}=\frac{\sqrt{5}}{\sqrt{2}}\Rightarrow |f-2g+\frac{5}{\sqrt{2}}|=\frac{5}{\sqrt{2}}\Rightarrow f-2g=-\frac{5}{\sqrt{2}}\pm \frac{5}{\sqrt{2}}\Rightarrow f-2g=0,f-2g=-5\sqrt{2}$

When $f=2g$, from equation $\left(2\right)$, we get
$\Rightarrow g=1,f=2\text{or}g=-1,f=-2$
Equation of the circle is
$x^2+y^2-2x-4y=0x^2+y^2+2x+4y=0$

When $f=2g-5\sqrt{2}$, from equation $\left(2\right)$, we get
$\Rightarrow g^2+(2g-5\sqrt{2}{)}^2=5\Rightarrow g^2-4\sqrt{2}g+9=0D=32-36<0$
No real solutions.

Hence, the required equation of the circle are
$x^2+y^2-2x-4y=0x^2+y^2+2x+4y=0$