The equation of circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$, where $c$ is

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

Let $B$ be the centre of the circle $x^{2}+y^{2}-2 x+4 y+1=0$ Let the tangents at two points $\mathrm{P}$ and $\mathrm{Q}$ on the circle intersect at the point $\mathrm{A}(3,1)$. Then $8.$ $\left(\frac{\text { area } \triangle \mathrm{APQ}}{\text { area } \triangle \mathrm{BPQ}}\right)$ is equal to .... .

If the line $3x - 4y = \lambda $ touches the circle ${x^2} + {y^2} - 4x - 8y - 5 = 0$, then $\lambda $ is equal to

If the tangent at the point $P$ on the circle ${x^2} + {y^2} + 6x + 6y = 2$ meets the straight line $5x - 2y + 6 = 0$ at a point $Q$ on the $y$- axis, then the length of $PQ$ is

Equation of a line through $(7, 4)$ and touching the circle, $x^2 + y^2 - 6x + 4y - 3 = 0$ is :