Question

If $(h,k)$ is the centre of the circle touches $y-\text{axis}$ at a distance of $12\text{units}$ from the origin and makes an intercept of $10\text{units}$ on $x-\text{axis}$, then the equation of circle for which $(h+k)$ is minimum, is

• A. $(x-13{)}^2+(y+12{)}^2=169$
• B. $(x+13{)}^2+(y-12{)}^2=169$
• C. $(x-13{)}^2+(y-12{)}^2=169$
• D. $(x+13{)}^2+(y+12{)}^2=169$

Answer 1

The correct option is D $(x+13{)}^2+(y+12{)}^2=169$

The centre of the circle is
$C\equiv (h,\pm 12)$

Assuming the point of intersection of the circle with $x-$ axis at $A$ and $B$, making a rough diagram, we get


In $△ABC$
$A{C}^2=A{M}^2+C{M}^2\Rightarrow r^2={12}^2+5^2=169\Rightarrow r=13$
Therefore the possible center of the circle are $(\pm 13,\pm 12)$
So, $h+k$ is minimum when $h=-13,k=-12$

$\therefore$ The required equation of the circle is
$(x+13{)}^2+(y+12{)}^2=169$