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 correct options are
A Area of quadrilateral $OACB=4$ sq. units
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 $(7,1)$

Coordinates of $O$ are $(5,3)$ and the radius $=2$

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
$5x+y-5(x+5)-3(y+1)+30=0\Rightarrow -2y+2=0\Rightarrow y=1$

$\therefore$ Coordinate of $C$ are $(7,1)$,
Length of tangent $CA=CB=2\text{units}$

$\therefore$ Area of $OACB=L.R=4\text{sq. units}$

Equation of $AB$ is
$x-y=4$ (radical axis)

Equation of the smallest circles is, when $AB$ is diameter,
$(x-7)(x-5)+(y-3)(y-1)=0\Rightarrow x^2+y^2-12x-4y+38=0$


Alternate Solution:
From above figure $OACB$ will be a square of side $2$ units
Hence required area $=4$ sq. units
Radical axis of the family of the circles passing through $A,B$ is
$AB:x-y=4$
The smallest possible circle have the $AB$ as diameter
$(x-7)(x-5)+(y-3)(y-1)=0$
i.e., $x^2+y^2-12x-4y+38=0$