If a circle with centre at (-1, 1) touches th | vedclass

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

Question

(A) The point of contact is the foot of the perpendicular from the centre $(-1, 1)$ to the line $x + 2y + 4 = 0$. Let the point of contact be $(h, k)$

Vedclass · Accepted Answer

$(-2, -1)$

Vedclass · Answer

(A) The point of contact is the foot of the perpendicular from the centre $(-1, 1)$ to the line $x + 2y + 4 = 0$. Let the point of contact be $(h, k)$. The line passing through the centre and perpendicular to the given line has a slope of $2$ (since the slope of $x + 2y + 4 = 0$ is $-1/2$). The equation of this perpendicular line is $y - 1 = 2(x + 1)$,which simplifies to $y = 2x + 3$. Substituting $y = 2x + 3$ into the given line equation $x + 2(2x + 3) + 4 = 0$: $x + 4x + 6 + 4 = 0$ $5x + 10 = 0$ $x = -2$. Substituting $x = -2$ into $y = 2x + 3$: $y = 2(-2) + 3 = -1$. Thus,the point of contact is $(-2, -1)$.

If a circle with centre at $(-1, 1)$ touches the line $x + 2y + 4 = 0$,then the coordinates of the point of contact are:

Similar Questions

Consider the circles $x^2 + (y - 1)^2 = 9$ and $(x - 1)^2 + y^2 = 25$. Which of the following is true?

Let $C$ be the circle with centre at $(1, 1)$ and radius $= 1$. If $T$ is the circle centred at $(0, y)$,passing through the origin and touching the circle $C$ externally,then the radius of $T$ is equal to:

The circle touching the coordinate axes with its centre lying on $x-2y-3=0$ is

Find the equation of the circumcircle of the triangle formed by the lines $y \pm \sqrt{3}x = 6$ and the $x$-axis.

If the line $y = mx + 1$ meets the circle $x^2 + y^2 + 3x = 0$ in two points equidistant from and on opposite sides of the $x$-axis,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module