$A$ circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is

The equation of the tangent to the curve given by $x=3 \cos \theta, y=3 \sin \theta$ at $\theta=\frac{\pi}{4}$ is

The point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ is

Let the lengths of intercepts on the $x$-axis and $y$-axis made by the circle $x^{2}+y^{2}+ax+2ay+c=0$ $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$,respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$ is equal to:

$A$ tangent is drawn to the circle $2x^{2} + 2y^{2} - 3x + 4y = 0$ at point $A$ and it meets the line $x + y = 3$ at $B(2, 1)$. Then,the length of $AB$ is equal to:

Abscissae of points on the curve $xy = (c + x)^2$,the normal at which cuts off numerically equal intercepts from the axes of coordinates is/are: