A, B and C are points on the circle with centre O. Q and R are points on the circle with centre P. CSR and BSQ are tangents on to both circles. AOB is a straight line.

1. State one pair of equal tangent lengths from $S$ to the left circle.

Since $SC$ and $SB$ are tangents from the same external point $S$ to the circle with centre $O$,

$SC = SB.$

2. State one pair of equal tangent lengths from $S$ to the right circle.

Since $SQ$ and $SR$ are tangents from the same external point $S$ to the circle with centre $P$,

$SQ = SR.$

3. Use the included angle and conclude congruence.

Lines $CSR$ and $BSQ$ intersect at $S$, so

$\angle CSQ = \angle BSR$

because they are vertically opposite angles.

Therefore, in $\triangle CQS$ and $\triangle BRS$,

$SC = SB, \quad SQ = SR, \quad \angle CSQ = \angle BSR.$

Hence,

$\triangle CQS \cong \triangle BRS$

by SAS.

1. Identify the right angle and use the angle sum of triangle $ABC$.

Since $AOB$ is a straight line through the centre $O$, $AB$ is a diameter.

Hence $\angle ACB=90^\circ$ (angle in a semicircle).

In $\triangle ABC$,
$\angle BAC+\angle ABC+\angle ACB=180^\circ$
$\angle BAC+y+90=180$
$\angle BAC=90-y^\circ$

1. Find the angle between the two tangents from $S$ to the left circle.

From part (b)(i), $\angle BAC=90-y^\circ$.

Since $\angle BOC$ is twice $\angle BAC$,
$\angle BOC=2(90-y)=180-2y^\circ.$

For the tangents $SB$ and $SC$, $OB\perp SB$ and $OC\perp SC$, so
$\angle BSC=180^\circ-\angle BOC=180-(180-2y)=2y^\circ.$

2. Relate this to the angle between the tangents to the right circle and find $\angle QPR$.

Since $B,S,Q$ are collinear and $C,S,R$ are collinear,
$\angle QSR=\angle BSC=2y^\circ$
by vertically opposite angles.

For the right circle, $PQ\perp SQ$ and $PR\perp SR$, so in quadrilateral $QPRS$,
$\angle QPR+\angle QSR=180^\circ.$

Hence
$\angle QPR=180^\circ-2y^\circ.$