The lengths of the tangents from the point (1 | vedclass

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

Question

(D) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c

Vedclass · Accepted Answer

$\frac{21}{4}$

Vedclass · Answer

(D) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.For the first circle $C_1: x^2+y^2+x+y-4=0$,the length of the tangent $L_1$ from $(1,2)$ is:$L_1 = \sqrt{1^2+2^2+1+2-4} = \sqrt{1+4+1+2-4} = \sqrt{4} = 2$.For the second circle $C_2: 3x^2+3y^2-x-y-k=0$,we first normalize it to $x^2+y^2-\frac{1}{3}x-\frac{1}{3}y-\frac{k}{3}=0$.The length of the tangent $L_2$ from $(1,2)$ is:$L_2 = \sqrt{1^2+2^2-\frac{1}{3}(1)-\frac{1}{3}(2)-\frac{k}{3}} = \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{k}{3}} = \sqrt{5-1-\frac{k}{3}} = \sqrt{4-\frac{k}{3}}$.Given the ratio $\frac{L_1}{L_2} = \frac{4}{3}$,we have:$\frac{2}{\sqrt{4-\frac{k}{3}}} = \frac{4}{3} \Rightarrow \frac{1}{\sqrt{4-\frac{k}{3}}} = \frac{2}{3}$.Squaring both sides:$\frac{1}{4-\frac{k}{3}} = \frac{4}{9} \Rightarrow 4-\frac{k}{3} = \frac{9}{4}$.$\frac{k}{3} = 4 - \frac{9}{4} = \frac{16-9}{4} = \frac{7}{4}$.$k = 3 	imes \frac{7}{4} = \frac{21}{4}$.

The lengths of the tangents from the point $(1,2)$ to the circles $x^2+y^2+x+y-4=0$ and $3x^2+3y^2-x-y-k=0$ are in the ratio $4:3$. Then the value of $k$ is:

Similar Questions

Let $S$ be the circumcircle of the triangle formed by the line $x-2y-4=0$ with the coordinate axes. If $P(-2, -4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $P$ and $Q$ is the least,then $PQ=$

Let $R$ be the region of the disc $x^2+y^2 \leq 1$ in the first quadrant. Then,the area of the largest possible circle contained in $R$ is

The length of the chord intercepted by the circle $x^2+y^2-4x+6y-12=0$ on the line $4x+3y+1=0$ is

If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}$,then the value of $r^{2}$ is equal to

For what values of $m$ does the circle $x^2 + y^2 = 4x + 8y + 5$ intersect the line $3x - 4y = m$ at two distinct points?

Vedclass Test Series

Exam Paper Generator

Online Exam Module