The equation of the pair of tangents drawn fr | vedclass

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

Question

(B) Let $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,we have $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$.The equation of the tangent $T$

Vedclass · Accepted Answer

$4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$

Vedclass · Answer

(B) Let $S = x^2 + y^2 - 2x + 4y = 0$. For the point $(x_1, y_1) = (0, 1)$,we have $S_1 = 0^2 + 1^2 - 2(0) + 4(1) = 5$.The equation of the tangent $T$ at $(x_1, y_1)$ is given by $x x_1 + y y_1 - (x + x_1) + 2(y + y_1) = 0$.Substituting $(0, 1)$,we get $T = x(0) + y(1) - (x + 0) + 2(y + 1) = -x + 3y + 2$.The equation of the pair of tangents is given by $SS_1 = T^2$.$5(x^2 + y^2 - 2x + 4y) = (-x + 3y + 2)^2$.$5x^2 + 5y^2 - 10x + 20y = x^2 + 9y^2 + 4 - 6xy - 4x + 12y$.Rearranging the terms,we get $4x^2 - 4y^2 + 6xy - 6x + 8y - 4 = 0$.

The equation of the pair of tangents drawn from the point $(0, 1)$ to the circle $x^2 + y^2 - 2x + 4y = 0$ is . . . . . .

Similar Questions

$y = mx$ is a chord of a circle of radius $a$. The diameter of the circle lies along the $x$-axis and one end of this chord is at the origin. The equation of the circle described on this chord as diameter is

The equation of the circle having the common chord of the circles $x^2+y^2-8x=0$ and $x^2+y^2-9=0$ as its diameter is

The equation of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2-5x-6y+4=0$ is

The perpendicular distance from the point $(1,2)$ to the common chord of the circles $x^2+y^2-2x+4y-4=0$ and $x^2+y^2+4x-6y-3=0$ is ........ units.

The equation of the chord of contact,if the tangents are drawn from the point $(5, -3)$ to the circle $x^2 + y^2 = 10$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module