Orthogonal Circles

We know that for intersection points, $C_{1}=C_{2}$ , and so

$\begin{array}{c}\\ (x-p)^{2}+(y-q)^{2}=x^{2}+y^{2}=r^{2} \Rightarrow \{ p^{2}+q^{2}=2px+2qy \: | \: x^{2}+y^{2}=r^{2} \} \quad (1) \end{array}$ .

For the circles to intersect orthogonally, $\frac{dy}{dx} \mid_{C_{1}} = -\frac{dx}{dy} \mid_{C_{2}}$ , so

$\begin{array}{c}\\ C_{1}: \quad 2x+2y \frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{x}{y} \\ C_{2}: \quad 2x-2p+ \left(2y-2q \right) \frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{x-p}{y-q} \end{array}$

And

$\begin{array}{c}\\ C_{1} \perp C_{2} \Rightarrow y^{2}-qy+x^{2}-px=0 \quad (2). \end{array}$

Since this will be happening at the intersection points of $C_{1}$ and $C_{2}$ , $x=x_{i}$ , $y=y_{i}$ and so condition I is true .

We can now combine $(1)$ and $(2)$ and remove the $x,y$ parameters, getting

$p^{2}+q^{2}=2r^{2} \: \: \: \: (3)$ .

The other two options can be checked from $(3)$ :

$\begin{array}{c}\\ q=a \Rightarrow 2p+0=4r\frac{dr}{dp} \Rightarrow \frac{dr}{dp}=\frac{p}{2r} \end{array}$ and so II is false .

$\begin{array}{c}\\ p+bq=0 \Rightarrow 2\left( b^{2}+1 \right) q=4r \frac{dr}{dq} \Rightarrow \frac{dr}{dq}=\frac{ b^{2}q+q}{2r} \end{array}$ and so III is true .

This is all of the options checked, so $\boxed{\text{I, III}}$ are true.