The equations of the normals to the circle ${x^2} + {y^2} - 8x - 2y + 12 = 0$ at the points whose ordinate is $-1,$ will be
$2x - y - 7 = 0,\,2x + y - 9 = 0$
$2x + y + 7 = 0,\,2x + y + 9 = 0$
$2x + y - 7 = 0,\,\,2x + y + 9 = 0$
$2x - y + 7 = 0,\,2x - y + 9 = 0$
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 .... .
The length of tangent from the point $(5, 1)$ to the circle ${x^2} + {y^2} + 6x - 4y - 3 = 0$, is
The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is
In the figure, $A B C D$ is a unit square. A circle is drawn with centre $O$ on the extended line $C D$ and passing through $A$. If the diagonal $A C$ is tangent to the circle, then the area of the shaded region is
If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to a circle, then the radius of the circle is