图中显示了一个以 $P(5, 3)$ 为圆心的圆 $C_1$。该圆与 x 轴、切线 OQ 以及另一条平行于 y 轴的直线 QR 相切。

1. Identify the centre and radius of the circle

The centre is $P(5,3)$.

Since the circle touches the x-axis, the radius is the vertical distance from $P$ to the x-axis:

$r=3$

2. Substitute the centre and radius into the equation of a circle

For a circle with centre $(a,b)$ and radius $r$,

$(x-a)^2+(y-b)^2=r^2$

Substituting $a=5$, $b=3$ and $r=3$,

$(x-5)^2+(y-3)^2=3^2$

$(x-5)^2+(y-3)^2=9$

1. Find the equation of the vertical mirror line QR

From 10(a)(i), $C_1$ has centre $(5,3)$ and radius $3$.

Since QR is the vertical tangent to the circle on the right,
$x=5+3=8.$

2. Reflect the centre and write the equation of the reflected circle

Reflecting $(5,3)$ in the line $x=8$ gives
$x'=8+(8-5)=11,$
so the reflected centre is $(11,3)$.

The radius is unchanged, so $r=3$.

Hence,
$(x-11)^2+(y-3)^2=3^2=9.$

1. Find $\tan\theta$ from the horizontal and vertical changes from $O$ to $P$

Since $O=(0,0)$ and $P=(5,3)$,

$\tan\theta=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{3}{5}.$

1. Represent the tangent line through the origin using an unknown gradient

Since $OQ$ passes through the origin, let its equation be
$y=mx.$

2. Substitute the line equation into the circle equation

From part (a)(i), the circle is
$(x-5)^2+(y-3)^2=9.$
Substitute $y=mx$:
$(x-5)^2+(mx-3)^2=9.$
Expand:
$x^2-10x+25+m^2x^2-6mx+9=9$
$(1+m^2)x^2+(-10-6m)x+25=0.$

3. Use the tangent condition to find the gradient

For the line to be a tangent, the quadratic has exactly one solution, so its discriminant is $0$:
$(-10-6m)^2-4(1+m^2)(25)=0.$
$100+120m+36m^2-100-100m^2=0$
$120m-64m^2=0$
$8m(15-8m)=0.$
So
$m=0 \quad \text{or} \quad m=\frac{15}{8}.$
The value $m=0$ gives the $x$-axis, not the tangent $OQ$. Hence
$m=\frac{15}{8}.$

4. Write the equation of the tangent line

Therefore the equation of $OQ$ is
$y=\frac{15}{8}x.$

1. Use the equation of the vertical line QR

From part (a)(ii), the line of reflection is QR, so
$x=8.$

2. Substitute into the tangent line equation to find the y-coordinate of Q

From part (b)(ii), the tangent line OQ is
$y=\frac{15}{8}x.$
At $Q$, $x=8$, so
$y=\frac{15}{8}(8)=15.$