The maximum area of a rectangle inscribed in | vedclass

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

Question

(C) The given equation of the circle is $(x+1)^{2}+(y-3)^{2}=64$. Comparing this with the standard form $(x-h)^{2}+(y-k)^{2}=r^{2}$,we get the radius

Vedclass · Accepted Answer

$128 	ext{ sq. units}$

Vedclass · Answer

(C) The given equation of the circle is $(x+1)^{2}+(y-3)^{2}=64$. Comparing this with the standard form $(x-h)^{2}+(y-k)^{2}=r^{2}$,we get the radius $r = \sqrt{64} = 8$. For a rectangle inscribed in a circle,the maximum area is achieved when the rectangle is a square. The diagonal of this square is equal to the diameter of the circle,which is $d = 2r = 2 	imes 8 = 16$. Let the side of the square be $a$. Then,by Pythagoras theorem,$a^{2} + a^{2} = d^{2}$. $2a^{2} = 16^{2} = 256$. $a^{2} = 128$. The area of the square is $a^{2} = 128 	ext{ sq. units}$.

The maximum area of a rectangle inscribed in the circle $(x+1)^{2}+(y-3)^{2}=64$ is

Similar Questions

Let $6$ and $8$ be the $X$ and $Y$-intercepts made by the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ respectively. If $gx+fy+1=0$ is a line passing through the point $(1, -1)$,then the radius of the circle $S=0$ is

If $\theta$ is the angle between the tangents from $(-1, 0)$ to the circle $x^2+y^2-5x+4y-2=0$,then $\theta$ is equal to

If a circle with centre at $(-1, 1)$ touches the line $x + 2y + 4 = 0$,then the coordinates of the point of contact are:

The equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2+y^2-6x-4y-12=0$ is

The equation of the chord of the circle $x^2+y^2-4x=0$ whose midpoint is $(1,0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module