Question

An equation of a circle through the origin,making an intercept.of $\sqrt{10}$ on the line $y=2x+\frac{5}{\sqrt{2}}$ , which subtends an angle of ${45}^{\circ }$ at the origin is

• 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)

The correct options are
B $x^2+y^2-2x-4y=0$
D $x^2+y^2+2x+4y=0$

Let an equation of the circle through the origin be$x^2+y^2+2gx+2fy=0$.

If $r$ is the radius of this circle, then $g^2+f^2=r^2$.
Let $PQ$ be the intercept of length $\sqrt{10}$ made by the line $y=2x+\frac{5}{\sqrt{2}}$ on this circle.

Since it is given that $PQ$ subtends an angle of ${45}^{\circ }$ at the origin (a point on the circle).

It subtends a right angle at the centre $(-g,-f)$ of the circle.
$\Rightarrow P{Q}^2=C{P}^2+C{Q}^2=2r^2$$\Rightarrow 10=2r^2\Rightarrow r=\sqrt{5}$

So that $g^2+f^2=5$ ....(1)

Let $CL$ be the perpendicular from $C$ on $PQ.$
Then $CL=LP=LQ=\frac{\sqrt{10}}{2}$$\Rightarrow \frac{-f+2g-\frac{5}{\sqrt{2}}}{\sqrt{1+4}}=\pm \frac{\sqrt{10}}{2}$
$\Rightarrow 2g-f=\pm \frac{5}{\sqrt{2}}+\frac{5}{\sqrt{2}}$$\Rightarrow 2g-f=0$ or $2g-f=5\sqrt{2}$

gives imaginary values of $g$ and $f$ from (1).

So $2g-f=0\Rightarrow f=2g$ and from (1) $g=\pm 1$ and the required equations are

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