Question

The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is

• A. $x^2+y^2-12x-16y=0$
• B. $x^2+y^2+12x+16y=0$
• C. $x^2+y^2+6x+8y=0$
• D. $x^2+y^2-6x-8y=0$

Answer 1

The correct option is D

$x^2+y^2-6x-8y=0$

The centre of the required circle is
$(\frac{6}{2},\frac{8}{2})=(3,4)$

The radius of the required circle is
$\sqrt{3^2+4^2}=\sqrt{25}=5$

Hence, the equation of the circle is as follows:

$(x-3{)}^2+(y-4{)}^2=5^2$

$\Rightarrow x^2+y^2-6x-8y=0$