The area of the circle (x - 2)^2 + (y - 3)^2 | vedclass

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

Question

(D) The equation of the circle is $(x - 2)^2 + (y - 3)^2 = 32$,which is $(x - 2)^2 + (y - 3)^2 = (4\sqrt{2})^2$. Here,the center of the circle is $(2,

Vedclass · Accepted Answer

$16 \pi$

Vedclass · Answer

(D) The equation of the circle is $(x - 2)^2 + (y - 3)^2 = 32$,which is $(x - 2)^2 + (y - 3)^2 = (4\sqrt{2})^2$. Here,the center of the circle is $(2, 3)$ and the radius $r = 4\sqrt{2}$. We check if the line $y = x + 1$ (or $x - y + 1 = 0$) passes through the center $(2, 3)$. Substituting $(2, 3)$ into the line equation: $2 - 3 + 1 = 0$. Since the center satisfies the equation of the line,the line is a diameter of the circle. $A$ diameter divides the circle into two equal areas. Therefore,the area of the circle lying below the line is half the total area of the circle. Area $= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4\sqrt{2})^2 = \frac{1}{2} \pi (32) = 16\pi$.

The area of the circle $(x - 2)^2 + (y - 3)^2 = 32$ that lies below the line $y = x + 1$ is:

Similar Questions

Area inside the parabola ${y^2} = 4ax$ between the lines $x = a$ and $x = 4a$ is equal to

The area bounded by the curve $y = -x^2$,the $x$-axis,$x = 1$,and $x = 4$ is:

The area of the region lying in the first quadrant and bounded by the curve $y = 4x^2$, the $Y$-axis, and the lines $y = 2$ and $y = 4$ is:

The area of the region bounded by the curve $x + 2y + 8 = 0$ and the lines $y = -3$ and $y = -1$ is . . . . . . sq. units.

$A$ curve satisfying the initial condition $y(1)= 0$ satisfies the differential equation $x \frac{dy}{dx}= y -x^2$. The area bounded by the curve and the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module