A square is inscribed in the circle x^2+y^2-1 | vedclass

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

Question

(D) The equation of the circle is $x^2+y^2-10x-6y+30=0$.Rewriting it as $(x-5)^2+(y-3)^2 = 25+9-30 = 4 = 2^2$.Thus,the center is $(5, 3)$ and the radi

Vedclass · Accepted Answer

$152$

Vedclass · Answer

(D) The equation of the circle is $x^2+y^2-10x-6y+30=0$.Rewriting it as $(x-5)^2+(y-3)^2 = 25+9-30 = 4 = 2^2$.Thus,the center is $(5, 3)$ and the radius $R = 2$.Let the sides of the square be parallel to $y=x+c$ and $x+y+d=0$.The distance from the center $(5, 3)$ to these lines must be equal to the distance from the center to the sides of the square,which is $R/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.For $y-x-c=0$: $\left|\frac{3-5-c}{\sqrt{1^2+(-1)^2}}\right| = \sqrt{2} \implies |c+2| = 2 \implies c=0$ or $c=-4$.For $x+y+d=0$: $\left|\frac{5+3+d}{\sqrt{1^2+1^2}}\right| = \sqrt{2} \implies |d+8| = 2 \implies d=-6$ or $d=-10$.The equations of the sides are $y=x$,$y=x-4$,$x+y=6$,and $x+y=10$.Solving these pairwise gives the vertices: $(5, 5), (3, 3), (5, 1), (7, 3)$.Calculating $\sum(x_i^2+y_i^2) = (25+25) + (9+9) + (25+1) + (49+9) = 50 + 18 + 26 + 58 = 152$.

$A$ square is inscribed in the circle $x^2+y^2-10x-6y+30=0$. One side of this square is parallel to $y=x+3$. If $(x_i, y_i)$ are the vertices of the square,then $\sum(x_i^2+y_i^2)$ is equal to:

Similar Questions

Find the equation of a circle that touches the $Y$-axis at a distance of $4$ units from the origin and cuts an intercept of $6$ units on the $X$-axis.

The length of the chord intercepted by the circle $x^2+y^2-6x+8y-5=0$ on the line $2x-y=5$ is equal to $L$ units. Find $L$.

The circle touching the coordinate axes with its centre lying on $x-2y-3=0$ is

$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

If one end of a diameter of the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is $(1, 0)$,then the other end of the diameter is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module