Circle $${x^2} + {y^2} - 4x - 8y - 5 = 0$$

Center $$=(2,4),$$ Radius $$ = \sqrt {4 + 16 + 5} = 5$$

If circle is intersecting line $$3x-4y=m,$$ at two distinct points.

$$ \Rightarrow $$ length of perpendicular from center to the line $$ < $$ radius

$$ \Rightarrow {{\left| {6 - 16 - m} \right|} \over 5} < 5 \Rightarrow \left| {10 + m} \right| < 25$$

$$ \Rightarrow - 25 < m + 10 < 25 \Rightarrow - 35 < m < 15$$

The given circles are

$${S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\therefore$$ Equation of common chord $$PQ$$ is $${S_1} - {S_2} = 0$$

$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$

$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is

$${S_1} + \lambda \,\,L = 0$$

$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda $$

$$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$

As it passes through $$\left( {1,1} \right),$$ therefore

$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$

$$ \Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},$$

which does not exist for $$p=-1$$

Let the center of the circle be $$(h, 2)$$

$$\therefore$$ Equation of circle is

$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Differentiating with respect to $$x,$$ we get

$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$

$$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$

Substituting in equation $$(1)$$ we get

$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$

$$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$

The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$



Center $$(-1,-2)$$

Let $$Q$$ $$\left( {\alpha ,\beta } \right)$$ be the point diametrically opposite to the point $$P(1,0),$$

then $${{1 + \alpha } \over 2} = - 1$$ and $${{0 + \beta } \over 2} = - 2$$

$$ \Rightarrow \alpha = - 3,\beta = - 4,$$ So, $$Q$$ is $$\left( { - 3, - 4} \right)$$