Let the circle x^{2}+y^{2}=4 intersect the x- | vedclass

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

Question

(A) The circle is $x^{2}+y^{2}=4$. The points of intersection with the $x$-axis are $A(2,0)$ and $B(-2,0)$.Let the point of intersection of $AQ$ and $

Vedclass · Accepted Answer

$x^{2}+y^{2}-4y-4=0$

Vedclass · Answer

(A) The circle is $x^{2}+y^{2}=4$. The points of intersection with the $x$-axis are $A(2,0)$ and $B(-2,0)$.Let the point of intersection of $AQ$ and $BP$ be $R(h,k)$.The slope of $BP$ is $m_{BP} = \frac{2 \sin \alpha - 0}{2 \cos \alpha - (-2)} = \frac{2 \sin \alpha}{2(\cos \alpha + 1)} = 	an \frac{\alpha}{2}$.The equation of line $BP$ is $y - 0 = 	an \frac{\alpha}{2} (x + 2)$.The slope of $AQ$ is $m_{AQ} = \frac{2 \sin \beta - 0}{2 \cos \beta - 2} = \frac{2 \sin \beta}{-2(1 - \cos \beta)} = -\cot \frac{\beta}{2}$.The equation of line $AQ$ is $y - 0 = -\cot \frac{\beta}{2} (x - 2)$.Given $\alpha - \beta = \frac{\pi}{2}$, so $\frac{\alpha}{2} - \frac{\beta}{2} = \frac{\pi}{4}$.From the equations, $	an \frac{\alpha}{2} = \frac{k}{h+2}$ and $\cot \frac{\beta}{2} = -\frac{k}{h-2}$.Using $	an(\frac{\alpha}{2} - \frac{\beta}{2}) = 	an \frac{\pi}{4} = 1$, we get $\frac{	an \frac{\alpha}{2} - 	an \frac{\beta}{2}}{1 + 	an \frac{\alpha}{2} 	an \frac{\beta}{2}} = 1$.Substituting the values, $\frac{\frac{k}{h+2} + \frac{h-2}{k}}{1 + (\frac{k}{h+2})(\frac{2-h}{k})} = 1$.This simplifies to $\frac{k^2 + h^2 - 4}{4k} = 1$, which gives $h^2 + k^2 - 4k - 4 = 0$.Thus, the locus is $x^2 + y^2 - 4y - 4 = 0$.

Similar Questions

If a point $P(x, y)$ moves such that the sum of the squares of its coordinates is equal to their product,then the locus of $P$ excluding the origin is

The equation of the locus of a point $(x, y)$ which is at a distance of $5$ units from a fixed point $(1, 4)$ and also at a distance of $5$ units from a fixed line $2x + 3y - 1 = 0$ is:

The set of all real values of $\lambda$ for which the point $P$ with coordinates $(\lambda, \lambda^2)$ does not lie inside the triangle formed by the lines $x - y = 0$, $x + y - 2 = 0$, and $x + 3 = 0$ is:

$A$ stick of length $r$ units slides with its ends on coordinate axes. Then the locus of the midpoint of the stick is a curve whose length is

The locus of the vertex $A$ of a triangle $ABC$,where the base $BC$ is fixed and the perimeter of the triangle is constant,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module