Question

The equation of circle which passes through the origin and cuts off intercepts $5$ and $6$ from the positive parts of the axes respectively, is ${(x-\frac{5}{2})}^2+(y-3)=\lambda$, where $\lambda$ is

• A. $\frac{61}{4}$
• B. $\frac{6}{4}$
• C. $\frac{1}{4}$
• D. $0$

Answer 1

The correct option is A $\frac{61}{4}$

From figure, we have
$OP=5,OQ=6$
and $OM=\frac{5}{2},CM=3$
Therefore, in $△OMC,O{C}^2=O{M}^2+M{C}^2$
$\Rightarrow O{C}^2={\left(\frac{5}{2}\right)}^2+(3{)}^2$
$\Rightarrow OC=\frac{\sqrt{61}}{2}$
Thus, the required circle has its centre $(\frac{5}{2},3)$ and radius $\frac{\sqrt{61}}{2}$.
So, its equation is ${(x-\frac{5}{2})}^2+(y-3{)}^2=\left(\frac{61}{4}\right)$.
Hence, $\lambda =\frac{61}{4}$