A square is inscribed in the circle x^{2} + y | vedclass

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

Question

(C) The given circle is $x^{2} + y^{2} - 2x + 4y - 93 = 0$.Completing the square,we get $(x - 1)^{2} + (y + 2)^{2} = 93 + 1 + 4 = 98$.The center is $(

Vedclass · Accepted Answer

$(-6, -9), (-6, 5), (8, -9), (8, 5)$

Vedclass · Answer

(C) The given circle is $x^{2} + y^{2} - 2x + 4y - 93 = 0$.Completing the square,we get $(x - 1)^{2} + (y + 2)^{2} = 93 + 1 + 4 = 98$.The center is $(1, -2)$ and the radius $R = \sqrt{98} = 7\sqrt{2}$.Let the side length of the square be $s$. Since the sides are parallel to the axes,the distance from the center $(1, -2)$ to each side is $s/2$.The distance from the center to a vertex is the radius $R = 7\sqrt{2}$.In a square with sides parallel to the axes,the distance from the center to a vertex is $\sqrt{(s/2)^{2} + (s/2)^{2}} = \frac{s}{\sqrt{2}}$.Equating this to the radius: $\frac{s}{\sqrt{2}} = 7\sqrt{2} \Rightarrow s = 14$.The sides are at distance $s/2 = 7$ from the center $(1, -2)$.The vertical sides are $x = 1 \pm 7$,i.e.,$x = -6$ and $x = 8$.The horizontal sides are $y = -2 \pm 7$,i.e.,$y = -9$ and $y = 5$.Thus,the vertices are $(-6, -9), (-6, 5), (8, -9), (8, 5)$.

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

Similar Questions

Which of the following pairs of straight lines intersect at right angles?

Power radiated by a black body at temperature $T_1$ is $P$ and it radiates maximum energy at a wavelength $\lambda_1$. If the temperature of the black body is changed from $T_1$ to $T_2$, it radiates maximum energy at a wavelength $\frac{\lambda_1}{2}$. The power radiated at $T_2$ is (in $P$)

Fluorosis disease is caused due to the reaction of ...... with excess of fluoride in the body.

$A$ ball is thrown on a lawn in such a way that it initially slides with a speed $v_0$ without rolling. It gradually picks up rotational motion. Find the speed of the ball at which there will be rolling without slipping.

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module