The point at which the normal to the circle $x^2 + y^2 + 4x + 6y - 39 = 0$ at the point $(2, 3)$ meets the circle again is:

Let $ABCD$ be a square of side length $1$. $A$ circle $C_{1}$ centered at $A$ with unit radius is drawn. Another circle $C_{2}$ which touches $C_{1}$ and is tangent to the lines $AD$ and $AB$ is also drawn. Let a tangent line from the point $C$ to the circle $C_{2}$ meet the side $AB$ at $E$. If the length of $EB$ is $\alpha+\sqrt{3} \beta,$ where $\alpha, \beta$ are integers,then $\alpha+\beta$ is equal to.........

The equation of the tangent at the point $\left( \frac{ab^2}{a^2 + b^2}, \frac{a^2b}{a^2 + b^2} \right)$ to the circle $x^2 + y^2 = \frac{a^2b^2}{a^2 + b^2}$ is

If the line $3x - 4y - k = 0 (k > 0)$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$ at $(a, b)$,then $k + a + b$ is equal to:

If $3x + y + k = 0$ is a tangent to the circle $x^2 + y^2 = 10$,then $k = . . . . . . $.

Let the equation of a circle be $x^{2}+y^{2}-6x-4y+9=0$. Then the line $4x+3y-8=0$ is a