Explanation

(b)
The centres of the two circles will lie on the line through $P(1,\ 2)$ perpendicular to the common tangent $4x + 3y = 10$. If $C_{1}$ and $C_{2}$ are the centres of these circles, then$PC_{1} = 5 = r_{1},PC_{2} = - 5 = r_{2}$. Also, $C_{1},C_{2}$ lie on the line$\frac{x - 1}{\cos\theta} = \frac{y - 2}{\sin\theta} = r$, where$\ \tan\theta = \frac{3}{4}$. When $r = r_{1}$ the coordinates of $C_{1}$ are $(5\cos\theta + 1,\ 5\sin\theta + 2)$ or (5, 5) as $\cos\theta = \frac{4}{5},\sin\theta = \frac{3}{5}$
When $r = r_{2}$, the coordinates of $C_{2}$ are $( - 3,\ - 1)$
The circle with centre $C_{1}(5,\ 5)$ and radius 5 touches both the coordinates axes and hence lies completely in the first quadrant
Therefore, the required circle is with centre $( - 3,\ - 1)$ and radius 5, so its equation is
$(x + 3)^{2} + (y + 1)^{2} = 5^{2}$
or $x^{2} + y^{2} + 6x + 2y - 15 = 0$
Since, the origin lies inside the circle, a portion of the circle lies in all the quadrants