Question

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

• A. The area of quadrilateral $OACB$ is $4$
• 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 The area of quadrilateral $OACB$ is $4$
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 $(7,1)$

The coordinates of $O$ are $(5,3)$ and the radius is $2.$
the equation of tangents at $A(7,3)$ is
$7x+3y-5(x+7)-3(y+3)+30=0\Rightarrow x=7$
The equation of tangent at $B(5,1)$ is
$5x+y-5(x+5)-3(y+1)+30=0\Rightarrow y=1$
Therefore,the coordinates of $C\text{are}(7,1)\text{Area of}OACB=4$
The equation of $AB$ is $x-y=4\text{(radical axis)}$
The equation of the smallest possible circle is
$(x-7)(x-5)+(y-3)(y-1)=0\Rightarrow x^2+y^2-12x-4y+38=0$