If P is a point such that the ratio of the sq | vedclass

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

Question

(B) Let the coordinates of $P$ be $(x, y)$.Given the equations of the circles:$S_1: x^2+y^2+2x-4y-20=0$$S_2: x^2+y^2-4x+2y-44=0$The square of the leng

Vedclass · Accepted Answer

$(-7,8)$

Vedclass · Answer

(B) Let the coordinates of $P$ be $(x, y)$.Given the equations of the circles:$S_1: x^2+y^2+2x-4y-20=0$$S_2: x^2+y^2-4x+2y-44=0$The square of the length of the tangent from $P(x, y)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $x^2+y^2+2gx+2fy+c$.Thus,the ratio of the squares of the lengths of the tangents is:$\frac{x^2+y^2+2x-4y-20}{x^2+y^2-4x+2y-44} = \frac{2}{3}$Cross-multiplying gives:$3(x^2+y^2+2x-4y-20) = 2(x^2+y^2-4x+2y-44)$$3x^2+3y^2+6x-12y-60 = 2x^2+2y^2-8x+4y-88$Rearranging the terms to one side:$x^2+y^2+14x-16y+28 = 0$This is the equation of a circle in the form $x^2+y^2+2gx+2fy+c=0$,where $2g=14$ and $2f=-16$.Thus,$g=7$ and $f=-8$.The centre of the circle is $(-g, -f) = (-7, 8)$.

If $P$ is a point such that the ratio of the square of the lengths of the tangents from $P$ to the circles $x^2+y^2+2x-4y-20=0$ and $x^2+y^2-4x+2y-44=0$ is $2:3$,then the locus of $P$ is a circle with centre :

Similar Questions

Let $A = (0, 4)$ and $B = (2 \cos \theta, 2 \sin \theta)$,for some $0 < \theta < \frac{\pi}{2}$. Let $P$ divide the line segment $AB$ in the ratio $2:3$ internally. The locus of $P$ is

The locus of the point of intersection of perpendicular tangents to the circle $x^{2}+y^{2}=16$ is

Any circle passing through the point of intersection of the lines $x + \sqrt{3}y = 1$ and $\sqrt{3}x - y = 2$ intersects these lines at points $P$ and $Q$. The angle subtended by the arc $PQ$ at its centre is- ............. $^o$

$A$ variable circle passes through a fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter passing through $A$ is:

The locus of the image of the point $(2, 3)$ in the line $(2x - 3y + 4) + k(x - 2y + 3) = 0, k \in R$ is a:

Vedclass Test Series

Exam Paper Generator

Online Exam Module