Question

Conisder the circle $x^2+y^2-10x-6y+30=0$. Let $O$ be the centre of the circle and tangent at $A(7,3)$ and $B(5,1)$ meet at $C$. Let $S=0$ represents family of circles passing through $A$ and $B$, then

• A. Area of quadrilateral $OACB=4$sq units
• B. The radical axis for the family of circles $S=0$ is $x+y=10$
• C. The smallest possible circle of the family $S=0$ is $x^2+y^2-12x-4y+38=0$
• D. The coordinates of point $C$ are $(7,1)$

Answer 1

Answer: (A), (C), (D)

The coordinates of point $C$ are $(7,1)$

$O\equiv (5,3)$,$r=\sqrt{5^2+3^2-30}=2$
Equation of tangent at $A(7,3)$ is
$7x+3y-5(x+7)-3(y+3)+30=0$
$\Rightarrow 2x-14=0$
$\Rightarrow x=7$

Equation of tangent at $B(5,1)$ is
$5x+y-5(x+5)-3(y+1)+30=0$
$\Rightarrow -2y+2=0$
$\Rightarrow y=1$
Coordinate of $C$ are $(7,1)$
$\therefore$ Area of $OACB=(7-5)\times (3-1)=4$

Equation of $AB:$
$y-1=\frac{3-1}{7-5}(x-5)$
$\Rightarrow x-y=4$ (radical axis)

Equation of the smallest circles passing through $A$ and $B$ is,
$(x-7)(x-5)+(y-3)(y-1)=0$
$\Rightarrow x^2+y^2-12x-4y+38=0$