Question

The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is
(a) x² + y² − 12x − 16y = 0
(b) x² + y² + 12x + 16y = 0
(c) x² + y² + 6x + 8y = 0
(d) x² + y² − 6x − 8y = 0

Answer 1

(d) x² + y² − 6x − 8y = 0

The centre of the required circle is $\left(\frac{6}{2},\frac{8}{2}\right)=\left(3,4\right)$.
The radius of the required circle is $\sqrt{3^2+4^2}=\sqrt{25}=5$.

Hence, the equation of the circle is as follows:
${\left(x-3\right)}^2+{\left(y-4\right)}^2=5^2$
$\Rightarrow$ $x^2+y^2-6x-8y=0$