Question

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

विकल्प

• x² + y² − 12x − 16y = 0

• x² + y² + 12x + 16y = 0

• x² + y² + 6x + 8y = 0

• x² + y² − 6x − 8y = 0

Answer 1

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:

\[\left( x - 3 \right)^2 + \left( y - 4 \right)^2 = 5^2\]

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