C.

24

B.

6

Let the equation of the circle is

x² + y² + 2gx +2fy +c =0

Thus circle touch the coordinate axes and lying in the first quadrant, then 

g² - c =0 and f² - c = 0

$$\begin{gathered} \mathrm{g} = \pm \sqrt{\mathrm{c}} , \mathrm{f} = \pm \sqrt{\mathrm{c}} \\ \mathrm{circle} \mathrm{lies} \mathrm{in} \mathrm{first} \mathrm{quadrant} \\ \therefore \mathrm{The} \mathrm{centre} \mathrm{is} \left(\sqrt{2}, \sqrt{2}\right) \\ \mathrm{If} \mathrm{the} \mathrm{line} 4\mathrm{x} + 3\mathrm{y} - 12 = 0 \mathrm{touch} \mathrm{the} \mathrm{circle}, \mathrm{then} \\ \sqrt{{\mathrm{f}}^2+ {\mathrm{g}}^2 - \mathrm{c}} = \frac{4\sqrt{\mathrm{c}} + 3\sqrt{\mathrm{c}} -12}{\sqrt{4^2 + 3^2}} \\ \Rightarrow \sqrt{\mathrm{c} + \mathrm{c}- \mathrm{c}} = \frac{7\sqrt{\mathrm{c}} - 12}{5} \\ \Rightarrow 5\sqrt{\mathrm{c}} = 7\sqrt{\mathrm{c}} - 12 \\ \Rightarrow 2\sqrt{\mathrm{c}} = 12 \Rightarrow \mathrm{c} = 36 \\ \mathrm{Radius} = \sqrt{{\mathrm{f}}^2+ {\mathrm{g}}^2 - \mathrm{c}} = \sqrt{\mathrm{c} + \mathrm{c}- \mathrm{c}} \sqrt{\mathrm{c}} = 6 \end{gathered}$$

A.

4x² - 4xy + y² - 8x + 4y + 4 = 0

Let P(x, y) be any point on the parabola

By defination of Parabola PM = PS

$$\begin{gathered} \frac{\mathrm{x} + 2\mathrm{y} - 1}{\sqrt{1+ 4}} = \sqrt{{\left(\mathrm{x} - 1\right)}^2 + {\mathrm{y}}^2} \\ \mathrm{On} \mathrm{squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get} \\ {\mathrm{x}}^2 + 4{\mathrm{y}}^2 +1 + 4\mathrm{xy} - 4\mathrm{y} -2\mathrm{x} \\ = 5\left({\mathrm{x}}^2 + 1 - 2\mathrm{x} + {\mathrm{y}}^2\right) \\ \Rightarrow 4{\mathrm{x}}^2 + {\mathrm{y}}^2 - 8\mathrm{x} + 4\mathrm{y} - 4\mathrm{xy} + 4 = 0 \end{gathered}$$

D.

- 20

$$\begin{gathered} \mathrm{Given} \mathrm{equation} \mathrm{of} \mathrm{circle} \mathrm{is} \\ {\mathrm{x}}^2 + {\mathrm{y}}^2 - 4\mathrm{x} - 2\mathrm{y} + \mathrm{c} = 0 \\ \mathrm{whose} \mathrm{centre} \mathrm{is} \mathrm{A}\left(2, 1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \mathrm{AP} = \sqrt{{\left(2 - 10\right)}^2 + {\left(1 - 7\right)}^2} \\ = \sqrt{{\left(- 8\right)}^2 + {\left(- 6\right)}^2} = \sqrt{64 + 36} = \sqrt{100} \\ = 10 \\ \therefore \mathrm{AQ} = \mathrm{AP} - \mathrm{PQ} = 10 - 5 = 5 \\ \mathrm{So}, \mathrm{Q} \mathrm{is} \mathrm{the} \mathrm{mid}-\mathrm{point} \mathrm{of} \mathrm{AP} = \left(\frac{10 + 2}{2}, \frac{7 + 1}{2}\right) = \left(6, 4\right) \\ \mathrm{Since}, \mathrm{Q} \mathrm{lies} \mathrm{on} \mathrm{a} \mathrm{circle} \\ \therefore 6^2 + 4^2 - 4\left(6\right) -2\left(4\right) + \mathrm{c} = 0 \\ \Rightarrow 36 + 16 - 24 - 8 + \mathrm{c} = 0 \\ \Rightarrow 20 + \mathrm{c} = 0 \\ \Rightarrow \mathrm{c} = - 20 \end{gathered}$$

C.

$(9, 4\sqrt{3})$

$$\begin{gathered} \mathrm{Given}, \mathrm{that} \mathrm{the} ∆\mathrm{SPM} \mathrm{is} \mathrm{equilateral} \\ \mathrm{Also}, \mathrm{given} \mathrm{parabola} \mathrm{is} {\mathrm{y}}^2 = 8\left(\mathrm{x} - 3\right) \mathrm{focus} \mathrm{of} \mathrm{this} \mathrm{parabola} \mathrm{is} \mathrm{S}(5, 0) \mathrm{and} \mathrm{vertex} \mathrm{A}(3, 0) \end{gathered}$$

Let coordinate of P (h + at², k + 2at) = P(3 + 2t², 4t)

Then, coordinate of M(- 5, 4t)

We know that the side of this  equilateral trianle is 4a = 4 x 2 = 8

Now, PS = 8

$$\begin{gathered} \sqrt{{\left(3 + 2{\mathrm{t}}^2 - 5\right)}^2 + {\left(4\mathrm{t}\right)}^2} = 8 \\ \Rightarrow \sqrt{{\left(2{\mathrm{t}}^2 - 2\right)}^2 + {\left(4\mathrm{t}\right)}^2} = 8 \\ \Rightarrow 2{\mathrm{t}}^2 + 2 = 8 \\ \Rightarrow 2{\mathrm{t}}^2 = 6 \\ \Rightarrow \mathrm{t} = \sqrt{3} \end{gathered}$$

$\therefore \mathrm{P}\left(3 + 2 \times 3, 4 \times \sqrt{3}\right) = \mathrm{P}\left(9, 4\sqrt{3}\right)$