ABCD is a square with side 16 units and A is | vedclass

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

Question

(B) The vertices of the square are $A(0,0)$,$B(16,0)$,$C(16,16)$,and $D(0,16)$.Since the circle circumscribes the square,the diagonal $AC$ is the diam

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The vertices of the square are $A(0,0)$,$B(16,0)$,$C(16,16)$,and $D(0,16)$.Since the circle circumscribes the square,the diagonal $AC$ is the diameter of the circle.The coordinates of $A$ are $(0,0)$ and $C$ are $(16,16)$.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 coordinates of $A$ and $C$:$(x-0)(x-16) + (y-0)(y-16) = 0$$x(x-16) + y(y-16) = 0$$x^2 - 16x + y^2 - 16y = 0$$x^2 + y^2 = 16(x+y)$Given the equation $x^2 + y^2 = 4k(x+y)$,we compare the two equations:$4k = 16$$k = 4$

$ABCD$ is a square with side $16$ units and $A$ is the origin. If the equation of the circle circumscribing the square $ABCD$ is $x^2+y^2=4k(x+y)$,then $k=$

Similar Questions

$(1, k)$ is a point on the circle passing through the points $(-1, 1), (0, -1)$ and $(1, 0)$. If $k \neq 0$,then $k =$

The parametric equations of the circle $x^2+y^2+2x-4y-4=0$ are

The diameters of a circle are along $2x + y - 7 = 0$ and $x + 3y - 11 = 0$. Then,the equation of this circle,which also passes through $(5, 7)$,is

If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8x-6y=0$,$4x^2+4y^2-4x-12y-186=0$,and $x^2+y^2-6x+6y-9=0$ respectively,then

The equation of circle $C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module