The locus of the midpoint of a chord of the c | vedclass

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

Question

(B) The angle subtended by the chord at the center of the circle $x^{2} + y^{2} = 4$ is $2 	imes 45^{\circ} = 90^{\circ}$.Let the midpoint of the cho

Vedclass · Accepted Answer

$x^{2} + y^{2} = 2$

Vedclass · Answer

(B) The angle subtended by the chord at the center of the circle $x^{2} + y^{2} = 4$ is $2 	imes 45^{\circ} = 90^{\circ}$.Let the midpoint of the chord $AB$ be $(h, k)$. The equation of the chord $AB$ is given by $T = S_1$,which is:$hx + ky - 4 = h^{2} + k^{2} - 4$$hx + ky = h^{2} + k^{2}$$\Rightarrow \frac{hx + ky}{h^{2} + k^{2}} = 1$Using this line to homogenize the circle equation $x^{2} + y^{2} - 4 = 0$,we get:$x^{2} + y^{2} - 4 \left( \frac{hx + ky}{h^{2} + k^{2}} ight)^{2} = 0$This represents the combined equation of lines $OA$ and $OB$. Since the angle at the center is $90^{\circ}$,the lines $OA$ and $OB$ are perpendicular.For perpendicular lines,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:$(1 - \frac{4h^{2}}{(h^{2} + k^{2})^{2}}) + (1 - \frac{4k^{2}}{(h^{2} + k^{2})^{2}}) = 0$$2 - \frac{4(h^{2} + k^{2})}{(h^{2} + k^{2})^{2}} = 0$$2 - \frac{4}{h^{2} + k^{2}} = 0$$h^{2} + k^{2} = 2$Thus,the locus is $x^{2} + y^{2} = 2$.

The locus of the midpoint of a chord of the circle $x^{2} + y^{2} = 4$ which subtends an angle of $45^{\circ}$ at the major arc of the circle is:

Similar Questions

$A$ point moves such that the square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is

$A$ circle touches the $X$-axis and also touches the circle with radius $2$ and center at $(0, 3)$. Find the locus of the center of the circle.

$A$ point moves such that the sum of the squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is always $6$. Then, its locus is

What does the equation $\sqrt{(x - 2)^2 + (y + 1)^2} + \sqrt{(x + 2)^2 + (y + 4)^2} = 5$ represent?

The locus of the centre of a circle touching both the coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module