If a tangent is drawn from the focus of the parabola $y^2 = 16x$ to the circle $(x - 6)^2 + y^2 = 2$,find the slope of this tangent.

$A(x_1, y_1)$ is the internal centre of similitude and $B(x_2, y_2)$ is the external centre of similitude of two circles $C_1$ and $C_2$ whose centres are $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ respectively. If $PA=3, AB=5, QB=2$,then the ratio of the radii of the two circles is:

The area of the triangle formed by joining the origin to the points of intersection of the line $x\sqrt{5} + 2y = 3\sqrt{5}$ and the circle $x^2 + y^2 = 10$ is

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then
$(A)$ $SP=2\sqrt{5}$
$(B)$ $SQ:QP=(\sqrt{5}+1):2$
$(C)$ the $x$-intercept of the normal to the parabola at $P$ is $6$
$(D)$ the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Let the parabola $y=x^2+px-3$ meet the coordinate axes at the points $P, Q$ and $R$. If the circle $C$ with center at $(-1,-1)$ passes through the points $P, Q$ and $R$,then the area of $\triangle PQR$ is:

The equation of a circle touching the parabola $y = x^2$ at the point $(1, 1)$ and passing through the point $(2, 2)$ is: