A rhombus is inscribed in the region common t | vedclass

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

Question

(D) The given circles are $C_1: x^2+y^2-4x-12=0$ and $C_2: x^2+y^2+4x-12=0$.Their centers are $A(-2, 0)$ and $B(2, 0)$, and both have radius $r = \sqr

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The given circles are $C_1: x^2+y^2-4x-12=0$ and $C_2: x^2+y^2+4x-12=0$.Their centers are $A(-2, 0)$ and $B(2, 0)$, and both have radius $r = \sqrt{2^2+0^2+12} = 4$.The common chord is obtained by subtracting the equations: $(x^2+y^2+4x-12) - (x^2+y^2-4x-12) = 0$, which gives $8x = 0$, or $x = 0$ (the $y$-axis).The intersection points $C$ and $D$ are found by substituting $x=0$ into $x^2+y^2-4x-12=0$, giving $y^2 = 12$, so $y = \pm 2\sqrt{3}$. Thus, $C(0, 2\sqrt{3})$ and $D(0, -2\sqrt{3})$.The diagonals of the rhombus are $AB$ (length $d_1 = 4$) and $CD$ (length $d_2 = 4\sqrt{3}$).The area of the rhombus is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 4 	imes 4\sqrt{3} = 8\sqrt{3}$ square units.

$A$ rhombus is inscribed in the region common to the two circles $x^2+y^2-4x-12=0$ and $x^2+y^2+4x-12=0$. If the line joining the centres of these circles and the common chord of them are the diagonals of this rhombus, then the area (in square units) of the rhombus is (in $\sqrt{3}$)

Similar Questions

If $L_1, L_2$ and $L_3$ are the chords of contact of the three points $(2,0), (1,-2)$ and $(4,4)$ respectively with respect to the circle $x^2+y^2=3$,then $L_1, L_2$ and $L_3$ are

If the circle $S \equiv x^2+y^2-4=0$ intersects another circle $S^{\prime}=0$ of radius $\frac{5 \sqrt{2}}{2}$ in such a manner that the common chord is of maximum length with slope equal to $\frac{1}{4}$,then the centre of $S^{\prime}=0$ is

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2-3x+y-10=0$ and $x^2+y^2-x+2y-20=0$ is

The circle $S \equiv x^2+y^2-2x-4y+1=0$ cuts the $y$-axis at $A, B$ $(OA > OB)$. If the radical axis of $S=0$ and $S^{\prime} \equiv x^2+y^2-4x-2y+4=0$ cuts the $y$-axis at $C$,then the ratio in which $C$ divides $AB$ is:

$A$ rhombus is inscribed in the region common to the two circles $x^2 + y^2 - 4x - 12 = 0$ and $x^2 + y^2 + 4x - 12 = 0$,with two of its vertices on the line joining the centers of the circles. The area of the rhombus is

Vedclass Test Series

Exam Paper Generator

Online Exam Module