The line $2x - y + 1 = 0$ is tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2y = 4$. The radius of the circle is

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the coordinate axes cut at right angles,then:

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.
$1.$ Equation of a common tangent with positive slope to the circle as well as to the hyperbola is:
$(A) 2x-\sqrt{5}y-20=0$
$(B) 2x-\sqrt{5}y+4=0$
$(C) 3x-4y+8=0$
$(D) 4x-3y+4=0$
$2.$ Equation of the circle with $AB$ as its diameter is:
$(A) x^2+y^2-12x+24=0$
$(B) x^2+y^2+12x+24=0$
$(C) x^2+y^2+24x-12=0$
$(D) x^2+y^2-24x-12=0$

For the given circle $2x^2 + 2y^2 = 5$ and the parabola $y^2 = 4\sqrt{5}x$:
Statement-$I$: The equation of the common tangent to these curves is $y = x + \sqrt{5}$.
Statement-$II$: If the line $y = mx + \frac{\sqrt{5}}{m} (m \neq 0)$ is a common tangent,then $m$ satisfies $m^4 - 3m^2 + 2 = 0$.

Two tangents $PQ$ and $PR$ are drawn to the circle $x^2 + y^2 - 2x - 4y - 20 = 0$ from the point $P(16, 7)$. If the centre of the circle is $C$,then the area of quadrilateral $PQCR$ is ............ $sq. \text{ units}$.

The coordinates of the point on the circle $x^2 + y^2 - 12x - 4y + 30 = 0$ which is farthest from the origin are: