The locus of the mid-points of chords of the | vedclass

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

Question

(B) Let $(h, k)$ be the coordinates of the mid-point of a chord of the circle $x^{2}+y^{2}=1$. The equation of the chord with mid-point $(h, k)$ is gi

Vedclass · Accepted Answer

$x^{2}+y^{2}=\frac{1}{2}$

Vedclass · Answer

(B) Let $(h, k)$ be the coordinates of the mid-point of a chord of the circle $x^{2}+y^{2}=1$. The equation of the chord with mid-point $(h, k)$ is given by $T=S_1$, which is $hx+ky = h^{2}+k^{2}$.The equation of the pair of lines joining the origin to the points of intersection of the circle and the chord is obtained by homogenizing the circle equation using the chord equation:$x^{2}+y^{2} = 1 \cdot \left(\frac{hx+ky}{h^{2}+k^{2}}ight)^{2}$$(h^{2}+k^{2})^{2}(x^{2}+y^{2}) = (hx+ky)^{2}$$(h^{2}+k^{2})^{2}(x^{2}+y^{2}) = h^{2}x^{2} + k^{2}y^{2} + 2hkxy$Since the chord subtends a right angle at the origin, the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:$(h^{2}+k^{2})^{2} - h^{2} + (h^{2}+k^{2})^{2} - k^{2} = 0$$2(h^{2}+k^{2})^{2} - (h^{2}+k^{2}) = 0$Since $h^{2}+k^{2} 
eq 0$, we have $2(h^{2}+k^{2}) = 1$, or $h^{2}+k^{2} = \frac{1}{2}$.Replacing $(h, k)$ with $(x, y)$, the locus is $x^{2}+y^{2} = \frac{1}{2}$.

The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$, which subtend a right angle at the origin, is

Similar Questions

The locus of the centers of the circles that are passing through the intersection of the circles $x^2+y^2=1$ and $x^2+y^2-2x+y=0$ is

$A$ cow is tied to a corner (vertex) of a regular hexagonal fenced area of side $a \ m$ by a rope of length $\frac{5a}{2} \ m$ in a grass field. (The cow cannot graze inside the fenced area). What is the maximum possible area of the grass field to which the cow has access to graze?

$A$ point $P$ moves in such a way that the ratio of its distance from two coplanar points is always a fixed number $(\lambda \ne 1)$. Then its locus is

If $A=(1,2), B=(2,1)$ and $P$ is a variable point satisfying the condition $|PA-PB|=3$,then the locus of $P$ is

The two points $A$ and $B$ in a plane are such that for all points $P$ lying on a circle satisfying $\frac{PA}{PB} = k$,then $k$ will not be equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module