Find the equations of the tangents to the cir | vedclass

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

Question

(A) The given circle is $x^2 + y^2 - 6x + 4y - 12 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = 2, c = -12$.The center is $(

Vedclass · Accepted Answer

$4x + 3y - 31 = 0, 4x + 3y + 19 = 0$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 - 6x + 4y - 12 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3, f = 2, c = -12$.The center is $(-g, -f) = (3, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 2^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$.The tangent is parallel to $4x + 3y + 5 = 0$,so its equation is of the form $4x + 3y + k = 0$.The perpendicular distance from the center $(3, -2)$ to the tangent line $4x + 3y + k = 0$ must be equal to the radius $r = 5$.Using the distance formula: $\frac{|4(3) + 3(-2) + k|}{\sqrt{4^2 + 3^2}} = 5$.$\frac{|12 - 6 + k|}{\sqrt{16 + 9}} = 5$.$\frac{|6 + k|}{5} = 5$.$|6 + k| = 25$.$6 + k = 25$ or $6 + k = -25$.$k = 19$ or $k = -31$.Thus,the equations of the tangents are $4x + 3y + 19 = 0$ and $4x + 3y - 31 = 0$.

Find the equations of the tangents to the circle $x^2 + y^2 - 6x + 4y = 12$ which are parallel to the line $4x + 3y + 5 = 0$.

Similar Questions

Let the lines $(2-i)z = (2+i)\bar{z}$ and $(2+i)z + (i-2)\bar{z} - 4i = 0$ (where $i^2 = -1$) be normal to a circle $C$. If the line $iz + \bar{z} + 1 + i = 0$ is tangent to this circle $C$,then its radius is

If the line $x+3y=0$ is the tangent at $(0,0)$ to the circle of radius $1$,then the centre of one such circle is

The equation of the tangent to the circle $x^2 + y^2 = a^2$ which is parallel to the line $y = mx + c$ is:

If the tangent and the normal at the point $(\sqrt{3}, 1)$ to the circle $x^2+y^2=4$,and the $X$-axis form a triangle,then the area (in sq. units) of this triangle is

The equation of the normal to the circle $2x^2 + 2y^2 - 2x - 5y + 3 = 0$ at the point $(1, 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module