If the sides of a rectangle are given by the | vedclass

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

Question

(C) The given equations of the sides are $x=-2, x=6, y=-2$ and $y=5$. The vertices of the rectangle are $A(-2, -2)$,$B(6, -2)$,$C(6, 5)$,and $D(-2, 5)

Vedclass · Accepted Answer

$x^2+y^2-4x-3y-22=0$

Vedclass · Answer

(C) The given equations of the sides are $x=-2, x=6, y=-2$ and $y=5$. The vertices of the rectangle are $A(-2, -2)$,$B(6, -2)$,$C(6, 5)$,and $D(-2, 5)$. The diagonal of the rectangle acts as the diameter of the circle. Taking diagonal $AC$ with endpoints $A(-2, -2)$ and $C(6, 5)$,the equation of the circle in diameter form is given by: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ $(x - (-2))(x - 6) + (y - (-2))(y - 5) = 0$ $(x + 2)(x - 6) + (y + 2)(y - 5) = 0$ $x^2 - 6x + 2x - 12 + y^2 - 5y + 2y - 10 = 0$ $x^2 + y^2 - 4x - 3y - 22 = 0$

If the sides of a rectangle are given by the equations $x=-2, x=6, y=-2, y=5$,then the equation of the circle,drawn on the diagonal of this rectangle as its diameter,is

Similar Questions

The equation of the circle having centre $(1, -2)$ and passing through the point of intersection of lines $3x + y = 14$ and $2x + 5y = 18$ is

Find the equation of the circle with center at $(0,0)$ and radius $r$.

The circle touching the $y$-axis at a distance $4$ units from the origin and cutting off an intercept $6$ from the $x$-axis is

$A$ circle is drawn touching the $X$-axis,with its centre at the point of reflection of $(m, n)$ on the line $y-x=0$. Then the equation of the circle is

Let the equation $x^{2}+y^{2}+px+(1-p)y+5=0$ represent circles of varying radius $r \in (0, 5]$. Then the number of elements in the set $S = \{q : q = p^{2} \text{ and } q \text{ is an integer}\}$ is ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module