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 $g = -1, f = 2, c = 1$.The radiu

Vedclass · Accepted Answer

$2\sqrt{2} 	ext{ units}$

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 $g = -1, f = 2, c = 1$.The radius $R = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (2)^2 - 1} = \sqrt{1 + 4 - 1} = \sqrt{4} = 2$.The distance $d$ from the center $(1, -2)$ to the point $P(-1, -4)$ is $d = \sqrt{(-1 - 1)^2 + (-4 - (-2))^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.The length of the chord of contact is given by $L = \frac{2R\sqrt{d^2 - R^2}}{d}$.Substituting the values: $L = \frac{2(2)\sqrt{(2\sqrt{2})^2 - 2^2}}{2\sqrt{2}} = \frac{4\sqrt{8 - 4}}{2\sqrt{2}} = \frac{4\sqrt{4}}{2\sqrt{2}} = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} 	ext{ units}$.

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

The length of the common chord of the circles $x^2+y^2+3x+5y+4=0$ and $x^2+y^2+5x+3y+4=0$ is

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

The equation of the pair of tangents at $(0,1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$ is

The length of the common chord of the circles ${x^2} + {y^2} + 5x + 7y + 9 = 0$ and ${x^2} + {y^2} + 7x + 5y + 9 = 0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module