The equations of the tangents to the circle x | vedclass

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

Question

(C) The equation of the 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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of the 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 $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{9+4+12} = \sqrt{25} = 5$. Any line parallel to $4x+3y+5=0$ is of the form $4x+3y+k=0$. The distance from the center $(3, -2)$ to the tangent line is equal to the radius: $\frac{|4(3)+3(-2)+k|}{\sqrt{4^{2}+3^{2}}} = 5$ $\frac{|12-6+k|}{5} = 5$ $|6+k| = 25$ $6+k = 25$ or $6+k = -25$ $k = 19$ or $k = -31$. Thus,the equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

The equations of the tangents to the circle $x^{2}+y^{2}-6x+4y=12$,which are parallel to the straight line $4x+3y+5=0$,are

Similar Questions

Two tangents to the circle $x^{2}+y^{2}=4$ at the points $A$ and $B$ meet at $M(-4,0)$. The area of the quadrilateral $MAOB$,where $O$ is the origin,is

The equation of the normal to the curve $x^{2}+y^{2}=r^{2}$ at the point $P(r \cos \theta, r \sin \theta)$ is:

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. Find the common tangent to the two circles.

If the line $3x - 4y = 1$ touches the circle $(x - 1)^2 + (y + 2)^2 = 4$ at $(\alpha, \beta)$,the values of $\alpha$ and $\beta$ are

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at $(1,-2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module