Tangents drawn from the origin to the circle | vedclass

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

Question

(C) The equation of the circle is $x^2 + y^2 - 2ax - 2by + b^2 = 0$.Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get th

Vedclass · Accepted Answer

$a^2 = b^2$

Vedclass · Answer

(C) The equation of the circle is $x^2 + y^2 - 2ax - 2by + b^2 = 0$.Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $(g, f) = (a, b)$ and radius $r = \sqrt{a^2 + b^2 - b^2} = \sqrt{a^2} = |a|$.For tangents drawn from the origin $(0, 0)$ to be perpendicular,the distance from the center $(a, b)$ to the origin must be equal to $\sqrt{2} 	imes r$.Distance from $(0, 0)$ to $(a, b)$ is $\sqrt{a^2 + b^2}$.So,$\sqrt{a^2 + b^2} = \sqrt{2} 	imes |a|$.Squaring both sides,we get $a^2 + b^2 = 2a^2$,which simplifies to $b^2 = a^2$ or $a^2 = b^2$.

Tangents drawn from the origin to the circle $x^2 + y^2 - 2ax - 2by + b^2 = 0$ are perpendicular to each other,if

Similar Questions

The angle of intersection between the curves $y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10,$ where $[x]$ denotes the greatest integer $\leq x,$ is

If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2}+y^{2}=4$ from the point $(3,2)$,then $m_{1}-m_{2}$ is equal to

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

The area (in sq units) of the triangle formed by the tangent,normal at $(1, \sqrt{3})$ to the circle $x^2+y^2=4$ and the $X$-axis,is

If $P(\frac{\pi}{4})$ and $Q(\frac{\pi}{3})$ are two points on the circle $x^2+y^2-2x-2y-1=0$,then the slope of the tangent to this circle which is parallel to the chord $PQ$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module