The area of the circle passing through the po | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the points be $A(1, 2)$, $B(5, 2)$, and $C(5, -2)$.Plotting these points, we observe that $AB$ is a horizontal line segment of length $4$ and

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let the points be $A(1, 2)$, $B(5, 2)$, and $C(5, -2)$.Plotting these points, we observe that $AB$ is a horizontal line segment of length $4$ and $BC$ is a vertical line segment of length $4$.Since $AB \perp BC$, the triangle $	riangle ABC$ is a right-angled triangle with the right angle at $B$.For a circle passing through the vertices of a right-angled triangle, the hypotenuse is the diameter of the circle.The hypotenuse is $AC = \sqrt{(5-1)^2 + (-2-2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.Thus, the diameter of the circle is $4\sqrt{2}$.The radius $r$ is half of the diameter, so $r = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.The area of the circle is $\pi r^2 = \pi(2\sqrt{2})^2 = \pi(8) = 8\pi$.

The area of the circle passing through the points $(5, 2), (5, -2),$ and $(1, 2)$ is (in $\pi$)

Similar Questions

The length of the tangent from the point $(5, 1)$ to the circle $x^2 + y^2 + 6x - 4y - 3 = 0$ is:

$A$ square is inscribed in the circle $x^2 + y^2 - 2x + 4y - 93 = 0$ with its sides parallel to the coordinate axes. The coordinates of its vertices are

Statement $1$: The only circle having radius $\sqrt{10}$ and a diameter along the line $2x + y = 5$ is $x^2 + y^2 - 6x + 2y = 0$.
Statement $2$: $2x + y = 5$ is a normal to the circle $x^2 + y^2 - 6x + 2y = 0$.

If the line $x + 2by + 7 = 0$ is a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$,then $b = $

Number of common tangents to the circles $x^2+y^2-6x-14y+48=0$ and $x^2+y^2-6x=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module