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

(A) The given circles are $C_1: x^2 + y^2 - 4x - 12 = 0$ with center $(2, 0)$ and radius $r = \sqrt{2^2 + 0^2 - (-12)} = \sqrt{16} = 4$.The second cir

Vedclass · Accepted Answer

$8\sqrt{3} 	ext{ sq. units}$

Vedclass · Answer

(A) The given circles are $C_1: x^2 + y^2 - 4x - 12 = 0$ with center $(2, 0)$ and radius $r = \sqrt{2^2 + 0^2 - (-12)} = \sqrt{16} = 4$.The second circle is $C_2: x^2 + y^2 + 4x - 12 = 0$ with center $(-2, 0)$ and radius $r = 4$.The common chord is obtained by subtracting the two equations: $(x^2 + y^2 - 4x - 12) - (x^2 + y^2 + 4x - 12) = 0$,which gives $-8x = 0$,so $x = 0$ (the $y$-axis).The intersection points of the circles with the $y$-axis are found by setting $x = 0$ in either equation: $y^2 - 12 = 0$,so $y = \pm 2\sqrt{3}$.The vertices of the rhombus are $(2, 0)$,$(-2, 0)$,$(0, 2\sqrt{3})$,and $(0, -2\sqrt{3})$.The diagonals of the rhombus are $d_1 = 2 - (-2) = 4$ and $d_2 = 2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3}$.The area of the rhombus is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 4 	imes 4\sqrt{3} = 8\sqrt{3} 	ext{ sq. units}$.

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

Similar Questions

$y = mx$ is a chord of a circle of radius $a$. The diameter of the circle lies along the $x$-axis and one end of this chord is at the origin. The equation of the circle described on this chord as diameter is

The length of the common chord of the circles $x^2 + y^2 + 2x + 3y + 1 = 0$ and $x^2 + y^2 + 4x + 3y + 2 = 0$ is

If the circle $C_1 : x^{2} + y^{2} = 16$ intersects another circle $C_2$ of radius $5$ such that the length of the common chord is maximum and its slope is $3/4$,then the coordinates of the center of $C_2$ are:

The radius of the circle,having centre at $(2, 1)$ and one of its chords as a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,is

Let $S$ be the focus of the parabola $y^2=8x$ and let $PQ$ be the common chord of the circle $x^2+y^2-2x-4y=0$ and the given parabola. The area of the triangle $PQS$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module