Question

Consider 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 the family of circles passing through $A$ and $B$, then

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

$x^2+y^2-10x-6y+30=0$
$\Rightarrow (x-5{)}^2+(y-3{)}^2=2^2$
$O,$ Centre $(5,3),$ Radius $=2$

$OACB$ is a square with side length $2$ units.
$\therefore$ Area of $OACB=4$
$S=0$ is the family of circles having common chord $AB.$
$A(7,3),B(5,1)$
$\therefore$ Equation of $AB$ is $y=x-4$
The smallest possible circle of family
$S=0$ has chord $AB$ as its diameter.
Required circle is
$(x-5)(x-7)+(y-1)(y-3)=0$
$\Rightarrow x^2+y^2-12x-4y+38=0$
Coordinates of C is $(7,1).$