The angle between the tangents to the circle | vedclass

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

Question

(D) The equation of the circle is $x^2 + y^2 = 169$,which has center $(0, 0)$ and radius $r = 13$.The equation of the tangent to the circle $x^2 + y^2

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) The equation of the circle is $x^2 + y^2 = 169$,which has center $(0, 0)$ and radius $r = 13$.The equation of the tangent to the circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$ is given by $xx_1 + yy_1 = r^2$.For the point $(5, 12)$,the tangent equation is $5x + 12y = 169$. The slope $m_1 = -\frac{5}{12}$.For the point $(12, -5)$,the tangent equation is $12x - 5y = 169$. The slope $m_2 = \frac{12}{5}$.Since $m_1 	imes m_2 = (-\frac{5}{12}) 	imes (\frac{12}{5}) = -1$,the tangents are perpendicular to each other.Therefore,the angle between the tangents is $90^o$.

The angle between the tangents to the circle $x^2 + y^2 = 169$ at the points $(5, 12)$ and $(12, -5)$ is ............. $^o$.

Similar Questions

If $3x + y + k = 0$ is a tangent to the circle $x^{2} + y^{2} = 10$,the values of $k$ are

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

If the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ cuts each of the three circles $x^2 + y^2 + 4x + 4y + 7 = 0$,$x^2 + y^2 - 4x + 4y + 7 = 0$,and $x^2 + y^2 - 4x - 4y + 7 = 0$ orthogonally,then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $S = 0$ is

The equations of the tangents to the curve $x^2 + y^2 = 10$, which are parallel to the line $2x + y - 1 = 0$, are:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module