Question

Circle $x^2+y^2=5$ and $x^2+y^2-3x+y=0$ intersect at point $A$ & $B$. Lines $OA$ and $OB$ are drawn through origin. Then

• A. Angle between $OA$ and $OB$ is ${90}^o$
• B. Length of $AB$ is $\sqrt{10}$ units
• C. Area of $\Delta AOB$ is $5$ sq. units
• D. Tangents at $A$ and $B$ to $x^2+y^2=5$ intersect at $(3,-1)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Length of $AB$ is $\sqrt{10}$ units
B Angle between $OA$ and $OB$ is ${90}^o$
D Tangents at $A$ and $B$ to $x^2+y^2=5$ intersect at $(3,-1)$

(A) Solving two equations of circle we get $A=(2,1)$ and $B=(1,-2)$.

So equation of $AB$ is $3x-y-5=0$.
On homogenising $AB$ with $x^2+y^2=5$,
Combined equation of $OA$ & $OB$ is
$x^2+y^2-5{\left(\frac{3x-y}{5}\right)}^2=0$
or $4x^2-4y^2-6xy=0$
$\therefore$ angle between $OA$ & $OB$ is ${90}^o$

(B) $L_{AB}=2\times AB=2\sqrt{(OA{)}^2-(OM{)}^2}$
$L_{AB}=2\sqrt{5-{\left(\frac{5}{\sqrt{10}}\right)}^2}=\sqrt{10}$

(C) Area of $\Delta AOB=\frac{1}{2}\left(AB\right)\left(OM\right)$
$=\frac{1}{2}\sqrt{10}.\frac{5}{\sqrt{10}}=\frac{5}{2}$

(D) Let tangents at $A$ & $B$ intersect at $(a,b)$
$\therefore$ $AB$ is $ax+by-5=0$ , $A=(-1,2),B=(-2,1)$
we get intersection point as$(3,-1)$
Ans: A,B,D