The product of the slopes of the common tange | vedclass

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

Question

(D) Let $S_1: x^2+y^2+2x-2y-2=0$. The center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - (-2)} = \sqrt{4} = 2$.Let $S_2: x^2+y^2-2x+2y+1=0$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let $S_1: x^2+y^2+2x-2y-2=0$. The center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - (-2)} = \sqrt{4} = 2$.Let $S_2: x^2+y^2-2x+2y+1=0$. The center $C_2 = (1, -1)$ and radius $r_2 = \sqrt{1^2 + (-1)^2 - 1} = \sqrt{1} = 1$.Since the circles are external to each other,the common tangents intersect at a point $P$ which divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio of their radii $r_1 : r_2 = 2 : 1$.Using the section formula for external division,the coordinates of $P$ are:$P = \left( \frac{2(1) - 1(-1)}{2-1}, \frac{2(-1) - 1(1)}{2-1} \right) = \left( \frac{3}{1}, \frac{-3}{1} \right) = (3, -3)$.Let the equation of the tangent line passing through $P(3, -3)$ be $y + 3 = m(x - 3)$,which simplifies to $mx - y - 3m - 3 = 0$.The perpendicular distance from the center $C_2(1, -1)$ to this line must equal the radius $r_2 = 1$.$\left| \frac{m(1) - (-1) - 3m - 3}{\sqrt{m^2 + (-1)^2}} \right| = 1$ $\Rightarrow \left| \frac{-2m - 2}{\sqrt{m^2 + 1}} \right| = 1$.Squaring both sides: $\frac{4(m+1)^2}{m^2+1} = 1 \Rightarrow 4(m^2 + 2m + 1) = m^2 + 1$.$4m^2 + 8m + 4 = m^2 + 1 \Rightarrow 3m^2 + 8m + 3 = 0$.The product of the slopes $m_1 m_2$ is given by the constant term divided by the coefficient of $m^2$ in the quadratic equation $3m^2 + 8m + 3 = 0$.Therefore,$m_1 m_2 = \frac{3}{3} = 1$.

The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y+1=0$ is:

Similar Questions

The equations of the tangents to the circle $x^2 + y^2 = 36$ which are inclined at an angle of $45^\circ$ to the $x$-axis are

If the line $y=2x+c$ is a tangent to the circle $x^2+y^2=5$,then a value of $c$ is

$A$ tangent is drawn to the circle $2x^{2} + 2y^{2} - 3x + 4y = 0$ at point $A$ and it meets the line $x + y = 3$ at $B(2, 1)$. Then,the length of $AB$ is equal to:

$3x + 4y - 43 = 0$ is a tangent to the circle $S \equiv x^2 + y^2 - 6x + 8y + k = 0$ at a point $P$. If $C$ is the centre of the circle and $Q$ is a point which divides $CP$ in the ratio $-1:2$,then the power of the point $Q$ with respect to the circle $S = 0$ is

$A$ circle $S=0$ with radius $\sqrt{2}$ touches the line $x+y-2=0$ at $(1,1)$. Then,the length of the tangent drawn from the point $(1,2)$ to $S=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module