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 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)

$x^2+y^2-10x-6y+30=0$

$\Rightarrow {(x-5)}^2+{(y-3)}^2={\left(2\right)}^2$

Centre $=(5,3),$ radius $=2$

$\Rightarrow oA=r=2$

$OB=r=2$

from figure,

$\Rightarrow AC=r=2$

$\Rightarrow BC=r=2$

$\Rightarrow C=(7,1)$

Area of $OACB=2\times 2=4$

Smallest circle of family $s=0$ will have $AB$ as diameter.

$\therefore$ Centre $=$ mid point of $AB=(6,2)$

radius $=\frac{AB}{2}=\frac{2\sqrt{2}}{2}=\sqrt{2}$

$\therefore$ Equation ${(x-6)}^2+{(y-2)}^2={\left(\sqrt{2}\right)}^2$

$\Rightarrow x^2-12x+36+y^2-4y+4=2$

$\Rightarrow x^2-12x+y^2-4y+38=0$

Radical axis of family of circles is $4B$

equation of $AB$ is

$\Rightarrow (y-1)=\frac{8-1}{7-5}(x-5)$

$-y-1=x-5$

$x-y-4=0$

$\therefore$ the answer is option A,C,D.