Aspire Faculty ID #11972 · Topic: NIMCET 2025 · Just now
NIMCET 2025

A circle with its center in the first quadrant touches both the coordinate axes and the line x-y-2=0. Then the area of the circle is

Solution

A circle touching both coordinate axes has center $(r, r)$ and radius $r$. 
It also touches the line $x - y - 2 = 0$. 
So the distance from $(r, r)$ to the line equals $r$: $\dfrac{|r - r - 2|}{\sqrt{1^{2} + (-1)^{2}}} = r$ 
$\dfrac{2}{\sqrt{2}} = r$ 
$r = \sqrt{2}$ 
Area of the circle: $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$

Previous 10 Questions — NIMCET 2025

Nearest first

Next 10 Questions — NIMCET 2025

Ascending by ID
Ask Your Question or Put Your Review.

loading...