Equation of the tangent to the circle, at the point $(1 , -1)$ whose centre is the point of intersection of the straight lines $x - y = 1$ and $2x + y= 3$ is
$x + 4y+ 3 = 0$
$3x - y- 4 = 0$
$x-3y-4 = 0$
$4x + y- 3 = 0$
The tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be -
If $OA$ and $OB$ be the tangents to the circle ${x^2} + {y^2} - 6x - 8y + 21 = 0$ drawn from the origin $O$, then $AB =$
A line meets the co-ordinate axes in $A\, \& \,B. \,A$ circle is circumscribed about the triangle $OAB.$ If $d_1\, \& \,d_2$ are the distances of the tangent to the circle at the origin $O$ from the points $A$ and $B$ respectively, the diameter of the circle is :
Equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which is perpendicular to the straight line $y = mx + c$ is
The two circles which passes through $(0,a)$ and $(0, - a)$ and touch the line $y = mx + c$ will intersect each other at right angle, if