Tangents OP and OQ are drawn from the origin | vedclass

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

Question

(B) The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the tangents are drawn from the origin $(0,0)$,the equation of the chord of c

Vedclass · Accepted Answer

$x^2 + y^2 + gx + fy = 0$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the tangents are drawn from the origin $(0,0)$,the equation of the chord of contact $PQ$ is given by $T = 0$,which is $gx + fy + c = 0$. Any circle passing through the intersection of the circle and the chord of contact is given by $(x^2 + y^2 + 2gx + 2fy + c) + \lambda(gx + fy + c) = 0$. Since this circle passes through the origin $(0,0)$,we substitute $x=0, y=0$: $(0^2 + 0^2 + 0 + 0 + c) + \lambda(0 + 0 + c) = 0 $ $\Rightarrow c + \lambda c = 0 $ $\Rightarrow \lambda = -1$ (assuming $c 
eq 0$). Substituting $\lambda = -1$ into the equation: $(x^2 + y^2 + 2gx + 2fy + c) - (gx + fy + c) = 0 \Rightarrow x^2 + y^2 + gx + fy = 0$.

Tangents $OP$ and $OQ$ are drawn from the origin $O$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Then,the equation of the circumcircle of the triangle $OPQ$ is

Similar Questions

The perpendicular distance from the point $(1,2)$ to the common chord of the circles $x^2+y^2-2x+4y-4=0$ and $x^2+y^2+4x-6y-3=0$ is ........ units.

Two circles centered at $(2,3)$ and $(5,6)$ intersect each other. If the radii are equal,the equation of the common chord is

If $L_1, L_2$ and $L_3$ are the chords of contact of the three points $(2,0), (1,-2)$ and $(4,4)$ respectively with respect to the circle $x^2+y^2=3$,then $L_1, L_2$ and $L_3$ are

If the circles $x^{2}+y^{2}-2x-2y-7=0$ and $x^{2}+y^{2}+4x+2y+k=0$ cut orthogonally,then the length of the common chord of the circles is

The common chord of the circles $x^2 + y^2 + 4x + 1 = 0$ and $x^2 + y^2 + 6x + 2y + 3 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module