The point of contact of the tangent to the ci | vedclass

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

Question

(B) The equation of the tangent to the circle $x^2 + y^2 = 5$ at the point $(x_1, y_1) = (1, -2)$ is given by $xx_1 + yy_1 = r^2$.Substituting the val

Vedclass · Accepted Answer

$(3, -1)$

Vedclass · Answer

(B) The equation of the tangent to the circle $x^2 + y^2 = 5$ at the point $(x_1, y_1) = (1, -2)$ is given by $xx_1 + yy_1 = r^2$.Substituting the values,we get $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$.To find the point of contact that also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$,we check which of the given options satisfies the equation of the tangent $x - 2y = 5$.For option $(B)$,$(3, -1)$: $3 - 2(-1) = 3 + 2 = 5$. This satisfies the equation.Thus,the point of contact is $(3, -1)$.

The point of contact of the tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ which also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$ is:

Similar Questions

If $\theta$ is the angle between the pair of tangents drawn from a point $P(x_1, y_1)$ to the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$,then which of the following is true?

Let a circle of radius $4$ and the ellipse $15x^2 + 19y^2 = 285$ be concentric. What is the angle that the common tangents make with the minor axis of the ellipse?

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y - 4 = 0$ is:

The two circles which pass through $(0, a)$ and $(0, -a)$ and touch the line $y = mx + c$ will intersect each other at a right angle,if

If $y=\sqrt{3}x+k_1$ and $y=\sqrt{3}x+k_2$ are two parallel tangents of a circle of radius $2 \text{ units}$,then $|k_1-k_2|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module