The equation of the circle whose diameter is | vedclass

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

Question

(A) The sides of the rectangle are $x=4, x=-2, y=5, y=-2$. The vertices of the rectangle are the intersection points of these lines: $(4, 5), (-2, 5),

Vedclass · Accepted Answer

$x^2+y^2-2x-3y-18=0$

Vedclass · Answer

(A) The sides of the rectangle are $x=4, x=-2, y=5, y=-2$. The vertices of the rectangle are the intersection points of these lines: $(4, 5), (-2, 5), (-2, -2),$ and $(4, -2)$. The diagonal of the rectangle connects opposite vertices,for example,$(4, 5)$ and $(-2, -2)$. The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$. Substituting the points $(4, 5)$ and $(-2, -2)$: $(x-4)(x+2) + (y-5)(y+2) = 0$ $x^2 + 2x - 4x - 8 + y^2 + 2y - 5y - 10 = 0$ $x^2 + y^2 - 2x - 3y - 18 = 0$.

The equation of the circle whose diameter is the diagonal of the rectangle with sides $x=4, x=-2, y=5, y=-2$ is:

Similar Questions

$A$ circle touches the axes at the points $(3, 0)$ and $(0, -3)$. The centre of the circle is

If the endpoints of the diameter of a circle are $(4, 3)$ and $(-12, -1)$,then the equation of the circle is . . . . .

The equation of the circle with centre $(2, 1)$ and touching the line $3x + 4y = 5$ is

If the segments of the straight lines $x+y=6$ and $x+2y=4$ are two diameters of a circle passing through $(6,2)$,then the equation of that circle is

The equation of a circle that passes through the origin and cuts off intercepts $-2$ and $3$ on the $X$-axis and $Y$-axis respectively is

Vedclass Test Series

Exam Paper Generator

Online Exam Module