Tangents are drawn from the point (-1, -4) to | vedclass

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

Question

(B) The equation of the circle is $x^2 + y^2 - 2x + 4y + 1 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = 1$.Ra

Vedclass · Accepted Answer

$2\sqrt{2}$ units

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 - 2x + 4y + 1 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = 1$.Radius $R = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - 1} = \sqrt{1 + 4 - 1} = \sqrt{4} = 2$.The length of the chord of contact from a point $(x_1, y_1)$ to a circle with radius $R$ is given by $L = \frac{2R\sqrt{S_1}}{R^2 + S_1}$,where $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.For point $(-1, -4)$,$S_1 = (-1)^2 + (-4)^2 - 2(-1) + 4(-4) + 1 = 1 + 16 + 2 - 16 + 1 = 4$.Since $S_1 = 4$ and $R = 2$,the length of the chord of contact is $L = \frac{2(2)\sqrt{4}}{2^2 + 4} = \frac{4(2)}{4 + 4} = \frac{8}{8} = 1$ unit. Wait,re-evaluating: The formula for the length of the chord of contact is $\frac{2R L_t}{\sqrt{R^2 + L_t^2}}$ where $L_t$ is the length of the tangent. $L_t = \sqrt{S_1} = \sqrt{4} = 2$.$L = \frac{2 	imes 2 	imes 2}{\sqrt{2^2 + 2^2}} = \frac{8}{\sqrt{8}} = \frac{8}{2\sqrt{2}} = 2\sqrt{2}$ units.

Tangents are drawn from the point $(-1, -4)$ to the circle $x^2 + y^2 - 2x + 4y + 1 = 0$. The length of the corresponding chord of contact is:

Similar Questions

Which of the following is a point on the common chord of the circles ${x^2 + y^2 + 2x - 3y + 6 = 0}$ and ${x^2 + y^2 + x - 8y - 13 = 0}$?

If the circle $x^2+y^2+2x+3y+1=0$ cuts another circle $x^2+y^2+4x+3y+2=0$ at points $A$ and $B$,then the equation of the circle with $AB$ as a diameter is

If the circles $x^2+y^2-2x-2y-7=0$ and $x^2+y^2+4x+2y+k=0$ cut orthogonally,then the length of their common chord is units.

The equation of the pair of tangents drawn from the origin to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is:

$A$ rhombus is inscribed in the region common to the two circles $x^2 + y^2 - 4x - 12 = 0$ and $x^2 + y^2 + 4x - 12 = 0$,with two of its vertices on the line joining the centers of the circles. The area of the rhombus is

Vedclass Test Series

Exam Paper Generator

Online Exam Module