Question 24

Mathematics Circle Medium

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

(A) \(2\pi\)
(B) \(\frac{\pi}{2}\)
(C) \(\pi\)
(D) \(4\pi\)
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

Correct Answer: A ($2\pi$)

The solution is found by using the geometric properties of the circle to determine its radius, and then calculating its area.

Step 1: Determine the Center and Radius of the Circle

The problem states that the circle has its center in the first quadrant and touches both the x-axis and the y-axis. This implies two things:

  • The distance from the center to the x-axis is equal to the radius, $r$.
  • The distance from the center to the y-axis is also equal to the radius, $r$.

Therefore, the coordinates of the center must be $(r, r)$ for some radius $r > 0$.

Step 2: Use the Tangent Line Condition to Find the Radius

A circle is tangent to a line if the perpendicular distance from its center to the line is equal to its radius.

  • Center of the circle: $(h, k) = (r, r)$
  • Tangent line: $x - y - 2 = 0$
  • Radius: $r$

The formula for the distance from a point $(h, k)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$.

Setting the distance equal to the radius, we get:

$r = \frac{|(1)(r) + (-1)(r) - 2|}{\sqrt{1^2 + (-1)^2}}$

$r = \frac{|r - r - 2|}{\sqrt{1 + 1}}$

$r = \frac{|-2|}{\sqrt{2}}$

$r = \frac{2}{\sqrt{2}} = \sqrt{2}$

So, the radius of the circle is $\sqrt{2}$.

Step 3: Calculate the Area of the Circle

The formula for the area of a circle is $Area = \pi r^2$.

Substituting the value of our calculated radius, $r = \sqrt{2}$:

$Area = \pi (\sqrt{2})^2 = 2\pi$

Therefore, the area of the circle is $2\pi$.