The equation of the pair of transverse common | vedclass

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

Question

(C) Given circles are $S_1 \equiv x^2 + y^2 + 2x + 2y + 1 = 0$ and $S_2 \equiv x^2 + y^2 - 2x - 2y + 1 = 0$. Centre of $S_1$ is $C_1 = (-1, -1)$ and r

Vedclass · Accepted Answer

$xy = 0$

Vedclass · Answer

(C) Given circles are $S_1 \equiv x^2 + y^2 + 2x + 2y + 1 = 0$ and $S_2 \equiv x^2 + y^2 - 2x - 2y + 1 = 0$. Centre of $S_1$ is $C_1 = (-1, -1)$ and radius $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = 1$. Centre of $S_2$ is $C_2 = (1, 1)$ and radius $r_2 = \sqrt{(1)^2 + (1)^2 - 1} = 1$. The distance between centres $C_1 C_2 = \sqrt{(1 - (-1))^2 + (1 - (-1))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. Since $C_1 C_2 > r_1 + r_2$ $(2\sqrt{2} > 2)$,the circles are separate. The transverse common tangents intersect at the internal centre of similitude $P$,which divides $C_1 C_2$ in the ratio $r_1 : r_2 = 1 : 1$. $P = \left( \frac{1(-1) + 1(1)}{1+1}, \frac{1(-1) + 1(1)}{1+1} \right) = (0, 0)$. The pair of tangents from $(0, 0)$ to $S_1$ is given by $T^2 = S_1 S_{11}$,where $T = x(0) + y(0) + 1(x+0) + 1(y+0) + 1 = x + y + 1$ and $S_{11} = 0^2 + 0^2 + 2(0) + 2(0) + 1 = 1$. Thus,$(x + y + 1)^2 = (x^2 + y^2 + 2x + 2y + 1)(1)$. $x^2 + y^2 + 1 + 2xy + 2x + 2y = x^2 + y^2 + 2x + 2y + 1$. $2xy = 0 \Rightarrow xy = 0$.

The equation of the pair of transverse common tangents drawn to the circles $x^2 + y^2 + 2x + 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ is

Similar Questions

Find the distance from the center of the circle $x^2 + y^2 = 2x$ to the common chord of the circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$.

If the midpoint of a chord of the circle $x^2 + y^2 + x - y - 1 = 0$ is $(1, 1)$,then the length of the chord is

From a point $P(-4, 0)$,two tangents are drawn to the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ touching the circle at $A$ and $B$. If the equation of the circle passing through $P, A$,and $B$ is $x^2 + y^2 + 2gx + 2fy + c = 0$,then $(g, f) =$

If a variable circle $S=0$ touches the line $y=x$ and passes through the point $(0,0)$,then the fixed point that lies on the common chord of the circles $x^2+y^2+6x+8y-7=0$ and $S=0$ is

If $OA$ and $OB$ are tangents from the origin to the circle $x^{2} + y^{2} - 6x - 8y + 21 = 0$,then $AB = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module