Question

The radical centre of three circles described on the three sides $4x-7y+10=0,x+y-5=0$ and $7x+4y-15=0$ of a triangle as diameters.

• A. $(1,2)$
• B. $(2,1)$
• C. $(1,-2)$
• D. $(1,1)$

Answer 1

The correct option is A $(1,2)$

Radical centre of three circles described on the sides of a triangle as diameters will be the orthocentre of the triangle.

Given sides are $4x-7y+10=0\dots \dots \left(1\right)$
$x+y-5=0\dots \dots \left(2\right)$
$7x+4y-15=0\dots \dots \left(3\right)$
Since lines $\left(1\right)$ and $\left(3\right)$ are perpendicular
So, the point of intersection of $\left(1\right)$ and $\left(3\right)$ is $(1,2)$ which is orthocentre of the triangle. Hence radical centre is $(1,2)$.