(a) To find the coordinates of $P$, which is on the $x$-axis, we set $y=0$ to obtain the equivalent equations

$$
\begin{aligned}
(x-6)^{2}+(0-8)^{2} & =100 \\
(x-6)^{2}+64 & =100 \\
(x-6)^{2} & =36 \\
x-6 & = \pm 6
\end{aligned}
$$

Therefore, $x-6=-6$ (or $x=0$ ) or $x-6=6$ (or $x=12$ ).

Since $O$ has coordinates $(0,0)$, then $P$ has coordinates $(12,0)$.

(b) Solution 1

The point, $Q$, with the maximum $y$-coordinate is vertically above the centre of the circle, $C$.

Since $C$ has coordinates $(6,8)$ and the radius of the circle is 10 , then $Q$ has coordinates $(6,8+10)$ or $(6,18)$.

Solution 2

Consider a point $(x, y)$ on the circle.

Since $(x-6)^{2}+(y-8)^{2}=100$, then $(y-8)^{2} \leq 100-(x-6)^{2}$.

Since $y>8$, we can maximize $y$ by maximizing $(y-8)^{2}$.

Since $(x-6)^{2} \geq 0$, then the maximum possible value of $(y-8)^{2}$ is 100 , when $(x-6)^{2}=0$ or $x=6$.

Since $y>8$, then $y-8=10$ or $y=18$.

This means that the point with the maximum $y$-coordinate is $Q(6,18)$.

(c) Solution 1

Since $\angle P Q R=90^{\circ}$, then a property of circles tells us that $P R$ is a diameter of the circle. Also, if $P R$ is a diameter of the circle, then $C$ (the centre of the circle) is its midpoint. Since $P$ has coordinates $(12,0)$ and $C$ has coordinates $(6,8)$, then if the coordinates of $R$ are $(a, b)$, we must have $6=\frac{1}{2}(12+a)$ (which gives $a=0$ ) and $8=\frac{1}{2}(b+0)$ (which gives $b=16)$.

Therefore, the coordinates of $R$ are $(0,16)$.

## Solution 2

Since $P$ has coordinates $(12,0)$ and $Q$ has coordinates $(6,18)$, then the slope of $P Q$ is $\frac{18-0}{6-12}=-3$.

Since $\angle P Q R=90^{\circ}$, then $P Q$ and $Q R$ are perpendicular.

Since the slope of $P Q$ is -3 , then the slope of $Q R$ is the negative reciprocal of -3 , which is $\frac{1}{3}$.

Since $Q$ has coordinates $(6,18)$, then the equation of the line through $Q$ and $R$ is $y-18=\frac{1}{3}(x-6)$ or $y=\frac{1}{3} x+16$.

We want to determine the second point of intersection of the line with equation $y=\frac{1}{3} x+16$ with the circle with equation $(x-6)^{2}+(y-8)^{2}=100$.

Substituting for $y$, we obtain

$$
\begin{aligned}
(x-6)^{2}+\left(\frac{1}{3} x+16-8\right)^{2} & =100 \\
9(x-6)^{2}+9\left(\frac{1}{3} x+8\right)^{2} & =900 \\
9\left(x^{2}-12 x+36\right)+(x+24)^{2} & =900 \\
9 x^{2}-108 x+324+x^{2}+48 x+576 & =900 \\
10 x^{2}-60 x & =0 \\
10 x(x-6) & =0
\end{aligned}
$$

Therefore, $x=6$ (which gives the point $Q$ ) or $x=0$.

When $x=0$, we obtain $y=16$, and so the coordinates of the point $R$ are $(0,16)$.

(d) Solution 1

Suppose that $X$ is a point on the circle for which $\angle P Q X=45^{\circ}$.

$P X$ is thus a chord of the circle that subtends an angle of $45^{\circ}$ on the circumference.

Therefore, $\angle P C X=2\left(45^{\circ}\right)=90^{\circ}$. That is, the central angle of chord $P X$ (which equals twice the subtended angle) is $90^{\circ}$ and so $X C$ is perpendicular to $P C$.

We now show two different ways to find the coordinates of the possible locations for $X$.

Method 1

Since $C$ has coordinates $(6,8)$ and $P$ has coordinates $(12,0)$, then to get from $C$ to $P$, we move right 6 units and down 8 units.

Since $X$ is on the circle, then $C X=C P$, because both are radii.

By using the same component distances, the length $C X$ will equal the length of $C P$. By interchanging the component distances and reversing the direction of one of these distances, we obtain a perpendicular slope. (In the diagram below, we can how how copies of the right-angled triangle with side lengths 6,8 and 10 fit together to form these right angles.)

Since $C X$ is perpendicular to $C P$, then possible locations for point $X$ can be found by moving right 8 units and up 6 units from $C$, and by moving left 8 units and down 6 units from $C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e1841c64bd565b3ef2b0g-09.jpg?height=412&width=463&top_left_y=2212&top_left_x=934)

Therefore, possible locations for $X$ are $(6+8,8+6)=(14,14)$ and $(6-8,8-6)=(-2,2)$, which are the possible coordinates of the points $S$ and $T$.

## Method 2

The slope of the line segment connecting $P(12,0)$ and $C(6,8)$ is $\frac{8-0}{6-12}=\frac{8}{-6}=-\frac{4}{3}$.

Since $X C$ is perpendicular to $P C$, then the slope of $X C$ is the negative reciprocal of $-\frac{4}{3}$, which equals $\frac{3}{4}$.

Therefore, the line through $X$ and $C$ has slope $\frac{3}{4}$ and passes through $C(6,8)$, and so has equation $y-8=\frac{3}{4}(x-6)$ or $y=\frac{3}{4} x-\frac{18}{4}+8$ or $y=\frac{3}{4} x+\frac{7}{2}$.

To find the possible locations of the point $X$ on the circumference of the circle, we substitute $y=\frac{3}{4} x+\frac{7}{2}$ into the equation of the circle to obtain the following equivalent equations:

$$
\begin{aligned}
(x-6)^{2}+\left(\frac{3}{4} x+\frac{7}{2}-8\right)^{2} & =100 \\
(x-6)^{2}+\left(\frac{3}{4} x-\frac{9}{2}\right)^{2} & =100 \\
(x-6)^{2}+\left(\frac{3}{4}(x-6)\right)^{2} & =100 \\
(x-6)^{2}\left(1+\left(\frac{3}{4}\right)^{2}\right) & =100 \\
(x-6)^{2}\left(\frac{25}{16}\right) & =100 \\
(x-6)^{2} & =64 \\
x-6 & = \pm 8
\end{aligned}
$$

Therefore, $x=8+6=14$ or $x=-8+6=-2$, which give $y=\frac{42}{4}+\frac{7}{2}=14$ and $y=\frac{-6}{4}+\frac{7}{2}=2$, respectively.

The two possible locations for $X$ are thus $(14,14)$ and $(-2,2)$, which are then the possible coordinates of the points $S$ and $T$.

## Solution 2

Consider the points $P(12,0)$ and $Q(6,18)$.

To move from $Q$ to $P$, we travel 6 units right and 18 units down.

Starting from $P$, if we move 18 units right and 6 units up to point $U$ with coordinates $(30,6)$, then $P U=P Q$ and $P U$ is perpendicular to $P Q$. (These facts follow from arguments similar to those in Solution 1, Method 1).

Similarly, starting from $P$, if we move 18 units left and 6 units down to point $V$ with coordinates $(-6,-6)$, then $P V=P Q$ and $P V$ is perpendicular to $P Q$.

Thus, $\triangle Q P U$ and $\triangle Q P V$ are right-angled and isosceles.

This means that $\angle P Q U=\angle P Q V=45^{\circ}$ (each is an angle in a right-angled, isosceles triangle), and so the points $S$ and $T$ where $Q U$ and $Q V$ respectively intersect the circle will give $\angle P Q S=\angle P Q T=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e1841c64bd565b3ef2b0g-10.jpg?height=439&width=615&top_left_y=2098&top_left_x=858)

The slope of the line segment joining $Q(6,18)$ and $U(30,6)$ is $\frac{18-6}{6-30}$ or $-\frac{1}{2}$.

Thus, the equation of the line through $Q$ and $U$ is $y-18=-\frac{1}{2}(x-6)$ or $y=-\frac{1}{2} x+21$.

Rearranging, we obtain $x=42-2 y$.

Substituting into the equation of the circle $(x-6)^{2}+(y-8)^{2}=100$, we obtain successively

$$
\begin{aligned}
(36-2 y)^{2}+(y-8)^{2} & =100 \\
4 y^{2}-144 y+1296+y^{2}-16 y+64 & =100 \\
5 y^{2}-160 y+1260 & =0 \\
y^{2}-32 y+252 & =0 \\
(y-18)(y-14) & =0
\end{aligned}
$$

The solution $y=18$ gives $x=42-2 y=6$, which are the coordinates of $Q$.

The solution $y=14$ gives $x=42-2 y=14$, which are the coordinates of the point $S$.

The slope of the line segment joining $Q(6,18)$ and $V(-6,-6)$ is $\frac{18-(-6)}{6-(-6)}$ or 2 .

Thus, the equation of the line through $Q$ and $V$ is $y-18=2(x-6)$ or $y=2 x+6$.

Substituting into the equation of the circle $(x-6)^{2}+(y-8)^{2}=100$, we obtain successively

$$
\begin{aligned}
(x-6)^{2}+(2 x-2)^{2} & =100 \\
x^{2}-12 x+36+4 x^{2}-8 x+4 & =100 \\
5 x^{2}-20 x-60 & =0 \\
x^{2}-4 x-12 & =0 \\
(x-6)(x+2) & =0
\end{aligned}
$$

The solution $x=6$ gives $y=2 x+6=18$, which are the coordinates of $Q$.

The solution $x=-2$ gives $y=2 x+6=2$, which are the coordintes of the point $T$.

Therefore, the coordinates of $S$ and $T$ are $(14,14)$ and $(-2,2)$.
In the diagram, the circle shown has centre $C(6,8)$, has radius 10 , and has equation

$$
(x-6)^{2}+(y-8)^{2}=100
$$

The circle passes through the origin $O(0,0)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_51c1a889111409851d36g-3.jpg?height=401&width=450&top_left_y=1792&top_left_x=1247)

(a) Determine the coordinates of the point $P$ where the circle intersects the $x$-axis.

(b) Determine, without justification, the coordinates of the point $Q$ on the circle that has the maximum $y$-coordinate among all points on the circle.

(c) Determine the coordinates of the point $R$ on the circle so that $\angle P Q R=90^{\circ}$.

(d) Determine the coordinates of the two distinct points $S$ and $T$ on the circle so that $\angle P Q S=\angle P Q T=45^{\circ}$.