The point (2, 3) is a limiting point of a coa | vedclass

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

Question

(A) The equation of a coaxial system of circles is given by $S + \lambda L = 0$,where $S = x^2 + y^2 - 9 = 0$ and $L$ is the radical axis.Since $(2, 3

Vedclass · Accepted Answer

$\left( \frac{18}{13}, \frac{27}{13} \right)$

Vedclass · Answer

(A) The equation of a coaxial system of circles is given by $S + \lambda L = 0$,where $S = x^2 + y^2 - 9 = 0$ and $L$ is the radical axis.Since $(2, 3)$ is a limiting point,the circle with radius $0$ passing through $(2, 3)$ is $(x-2)^2 + (y-3)^2 = 0$,which is $x^2 + y^2 - 4x - 6y + 13 = 0$.The radical axis is $S_1 - S_2 = 0$,which gives $(x^2 + y^2 - 9) - (x^2 + y^2 - 4x - 6y + 13) = 0$,simplifying to $4x + 6y - 22 = 0$ or $2x + 3y - 11 = 0$.The family of coaxial circles is $(x^2 + y^2 - 9) + k(2x + 3y - 11) = 0$.The center is $(-k, -\frac{3k}{2})$ and the radius squared is $r^2 = k^2 + \frac{9k^2}{4} + 9 + 11k = \frac{13k^2 + 44k + 36}{4}$.For limiting points,$r^2 = 0$,so $13k^2 + 44k + 36 = 0$.Solving for $k$,we get $(13k + 18)(k + 2) = 0$,so $k = -2$ or $k = -\frac{18}{13}$.For $k = -2$,the center is $(2, 3)$.For $k = -\frac{18}{13}$,the center is $(\frac{18}{13}, \frac{3}{2} 	imes \frac{18}{13}) = (\frac{18}{13}, \frac{27}{13})$.

The point $(2, 3)$ is a limiting point of a coaxial system of circles of which $x^2 + y^2 = 9$ is a member. The coordinates of the other limiting point are given by

Similar Questions

The coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

The centre of the circle which intersects the circle $x^2+y^2-2x-2y-2=0$ orthogonally,passes through the point $(2,0)$,and touches the $X$-axis is:

The equation of the circle which passes through the origin,has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally,is

The circles $x^2+y^2+2x+3y-7=0$ and $x^2+y^2+4x-7y+5=0$ intersect at the points $A$ and $B$. The equation of the circle,having $\overline{AB}$ as a diameter is

$A$ circle $S$ passes through the points of intersection of the circles $x^2+y^2-2x+2y-2=0$ and $x^2+y^2+2x-2y+1=0$. If the centre of this circle $S$ lies on the line $x-y+6=0$,then the radius of the circle $S$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module