Sec 3 A Math: Coordinate Geometry of Circles, practice questions & worked solutions

$$\begin{aligned} (x-1)^2+(y-3)^2 &= 5 \text{------(1)}\\ y-3x &= 5\\ y &= 3x+5 \text{-----------(2)} \end{aligned}$$ Subst (2) to (1): $$\begin{aligned} (x-1)^2+(3x+5-3)^2 &= 5\\ x^2-2x+1+9x^2+12x+4 &= 5\\ 10x^2+10x &= 0\\ 10x(x+1) &= 0\\ x=0 \;&\text{or}\; x=-1\\ y=5 \;&\text{or}\; y=2 \end{aligned}$$ the coordinates are $(0,5)$ and $(-1,2)$.

(a)

$$\begin{aligned} x^2+y^2-8x-10y+16 &= 0\\ (x-4)^2-4^2+(y-5)^2-5^2+16 &= 0\\ (x-4)^2+(y-5)^2 &= 25\\ C(4,5), &\text{radius}=5\ \text{units} \end{aligned}$$

(b)

$$\begin{aligned} \text{Gradient of } BC &= \frac{9-5}{7-4} = \frac{4}{3}\\ \text{Gradient of tangent at } B &= -\frac{3}{4}\\ \text{Equation of tangent at } B:\ y-9 &= -\frac{3}{4}(x-7)\\ y &= -\frac{3}{4}x+\frac{57}{4} \end{aligned}$$

(c)

$P$ is vertically below centre $C$.

$x$-coordinate of $P = x$-coordinate of $C = 4$.

$y$-coordinate of $P = 5-\text{radius}=5-5=0$.

$\therefore P(4,0)$ lies on the $x$-axis.

(a)

$$\begin{aligned} y + 3 &= \frac{4}{3}(x + 1)\\ y &= \frac{4}{3}x - \frac{5}{3}\\ 3y - 4x &= -5 \cdots\text{(i)}\\ 4y + 3x &= 10 \cdots\text{(ii)}\\ 9y - 12x &= -15 \cdots\text{(i)}\\ 16y + 12x &= 40 \cdots\text{(ii)}\\ 25y &= 25\\ y &= 1\\ x &= 2 \end{aligned}$$ Centre $(2, 1)$ and radius $\sqrt{(2 + 1)^2 + (1 + 3)^2} = 5$

Equation of circle is $(x - 2)^2 + (y - 1)^2 = 25$

(b)

radius of $C_2 = 3$ $$\begin{aligned} (x + 1)^2 + (y + 3)^2 &= 3^2\\ x^2 + 2x + 1 + y^2 + 6y + 9 &= 9\\ x^2 + y^2 + 2x + 6y + 1 &= 0 \end{aligned}$$

(a)

Let the $y$-coordinate of $P$ be $a$, $$\begin{aligned} \frac{4 - a}{5 - (-1)} &= 1\\ 4 - a &= 6\\ a &= -2 \end{aligned}$$ Coordinates of $P$ is $(-1, -2)$.

(b)

(i) $$\begin{aligned} \text{Coordinate } C &= (5 - 2, 4 - 2)\\ &= (3, 2) \end{aligned}$$

(ii) $$\begin{aligned} \text{Distance } BC &= \sqrt{(5 - 3)^2 + (4 - 2)^2}\\ &= \sqrt{8} \ \text{units} \end{aligned}$$ Equation of circle is $(x - 3)^2 + (y - 2)^2 = 8$.

(c)

Gradient of $CD = -1$

Equation of $CD$ is $$\begin{aligned} y - 2 &= -1(x - 3)\\ y &= -x + 5 \end{aligned}$$ Subst. $y = -x + 5$ into $(x - 3)^2 + (y - 2)^2 = 8$, $$\begin{aligned} (x - 3)^2 + (-x + 5 - 2)^2 &= 8\\ x^2 - 6x + 9 + x^2 - 6x + 9 &= 8\\ 2x^2 - 12x + 10 &= 0\\ x^2 - 6x + 5 &= 0\\ (x - 1)(x - 5) &= 0\\ x = 1 \ \text{or} \ x &= 5\ (\text{NA}) \end{aligned}$$ Subst. $x = 1$ into $y = -x + 5$, $$\begin{aligned} y &= -1 + 5\\ y &= 4\\ D&(1, 4) \end{aligned}$$

(a)

$$\begin{aligned} x^2 - 6x + y^2 + 4y &= 12\\ (x - 3)^2 + (y + 2)^2 &= 12 + 3^2 + 2^2\\ (x - 3)^2 + (y + 2)^2 &= 5^2 \end{aligned}$$ Radius $= 5$ units

Centre $= (3, -2)$

(b)

$$\begin{aligned} \text{Gradient of normal} &= \frac{-2 - (-5)}{3 - 7} = -\frac{3}{4}\\ \text{Gradient of tangent} &= \frac{4}{3} \end{aligned}$$ Eqn of tangent: $$\begin{aligned} y + 5 &= \frac{4}{3}(x - 7)\\ y &= \frac{4}{3}x - \frac{43}{3} \end{aligned}$$

(c)

$$\begin{aligned} \text{Distance between centres of circle} &= \sqrt{(3 + 8)^2 + (-2 - 4)^2} = \sqrt{157}\\ \text{Shortest distance} &= \sqrt{157} - 7 - 5 = 0.530\ \text{cm (3 s.f.)} \end{aligned}$$

(a)

$$\begin{aligned} x^2 + y^2 - 10x + 6y + 9 &= 0\\ x^2 - 10x + 25 + y^2 + 6y + 9 - 25 - 9 + 9 &= 0\\ (x - 5)^2 + (y + 3)^2 &= 5^2 \end{aligned}$$ Centre of Circle $= (5, -3)$

Radius $= 5$ units

(b)

For using the correct method to find the perpendicular distance from the centre of the circle to the $y$-axis. Distance from centre $(5, -3)$ to $y$-axis $(x = 0) = 5$

Distance equals radius, so $y$-axis is a tangent.

(c)

$$\begin{aligned} \text{Gradient of } CP &= \frac{0 - (-3)}{1 - 5} = -\frac{3}{4}\\ \text{Gradient of tangent} &= \frac{4}{3}\\ y &= \frac{4}{3}x - \frac{4}{3} \end{aligned}$$

(d)

$$\begin{aligned} \text{LHS} &= (3 - 5)^2 + (-7 + 3)^2 = 20 \end{aligned}$$ Since $20 < 25$, $(3, -7)$ lies inside the circle.

(a)

$$\begin{aligned} m_{AB} &= 1 \end{aligned}$$ Equation of $AB$: $$\begin{aligned} y + 2 &= x + 4\\ y &= x + 2 \text{----(1)}\\ y &= -x + 4 \text{----(2)}\\ (1) = (2): \ x + 2 &= -x + 4\\ 2x &= 2\\ x &= 1\\ y &= 3 \end{aligned}$$ Coordinates of $A = (1, 3)$

(b)

$$\begin{aligned} \text{Radius} &= \sqrt{(1 + 4)^2 + (3 + 2)^2}\\ &= \sqrt{50}\ \text{units} \end{aligned}$$ Equation of circle: $$(x + 4)^2 + (y + 2)^2 = 50$$

(c)

Let the coordinates of the other end of the diameter passes through $A = C(x, y)$.
Since $AC$ is the diameter, $B$ is the midpoint of $AC$. $$\begin{aligned} \left(\frac{x + 1}{2},\ \frac{y + 3}{2}\right) &= (-4,\ -2)\\ \frac{x + 1}{2} = -4, \frac{y + 3}{2} &= -2\\ x = -9, y &= -7 \end{aligned}$$ Equation of another tangent which is parallel to $y = -x + 4$: $$\begin{aligned} y + 7 &= -1(x + 9)\\ y &= -x - 16 \end{aligned}$$

(a)

$$\begin{aligned} P&(-13, 11)\\ Q&(\ 1,\ 9)\\ R&(-1, -5) \end{aligned}$$ $$\begin{array}{c|c} \text{Gradient of } PQ & \text{Gradient of } QR\\[2pt] m_{PQ} & m_{QR}\\[4pt] = \dfrac{11 - 9}{-13 - 1} & = \dfrac{9 - (-5)}{1 - (-1)}\\[6pt] = -\dfrac{1}{7} & = 7 \end{array}$$ Since $$m_{PQ} \times m_{QR} = -\frac{1}{7} \times 7 = -1$$ $$\begin{aligned} &\Rightarrow PQ \text{ is perpendicular to } QR\\ &\Rightarrow \angle PQR = 90^\circ (\textbf{right angle in semicircle})\\ &\Rightarrow PR \text{ is the diameter of circle.} \end{aligned}$$

Or

$$\begin{aligned} PQ^2 &= (-13 - 1)^2 + (11 - 9)^2\\ &= 200\\ QR^2 &= (1 - (-1))^2 + (9 - (-5))^2\\ &= 200\\ PR^2 &= (-13 + 1)^2 + (11 + 5)^2\\ &= 400 \end{aligned}$$ Since $PR^2 = PQ^2 + QR^2 = 400$, by converse of Pythagoras’ Theorem, $\angle PQR = 90^\circ$.  (right angle in semicircle) $$\Rightarrow PR \text{ is the diameter of circle.}$$

Or

Midpoint of $PR$, $$\begin{aligned} M &= \left(\frac{-13 - 1}{2},\ \frac{11 - 5}{2}\right)\\ &= (-7,\ 3) \ \cdots [1]\\ PM &= \sqrt{(-13 + 7)^2 + (11 - 3)^2}\\ &= 10\\ QM &= \sqrt{(1 + 7)^2 + (9 - 3)^2}\\ &= 10\\ RM &= \sqrt{(-1 + 7)^2 + (-5 - 3)^2} = 10\\ \therefore PM &= QM = RM = 10 \ \cdots [2] \end{aligned}$$ From [1] and [2], midpoint of chord $PR$ is equidistant to $P$, $Q$ & $R$ $$\Rightarrow PR \text{ is the diameter of circle.}$$

(b)

As centre is midpoint of diameter, $$\begin{aligned} \text{centre} &= \left(\frac{-13 - 1}{2},\ \frac{11 - 5}{2}\right)\\ &= (-7, 3)\\ \text{radius} &= \frac{1}{2}PR\\ &= \frac{1}{2}\sqrt{(-13 - (-1))^2 + (11 - (-5))^2}\\ &= \frac{1}{2}\sqrt{400}\\ &= 10 \end{aligned}$$ Equation of circle $$\begin{aligned} (x - (-7))^2 + (y - 3)^2 &= 10^2\\ x^2 + 14x + 49 + y^2 - 6y + 9 &= 100\\ x^2 + y^2 + 14x - 6y - 42 &= 0 \end{aligned}$$

(c)

Let $T(x, y)$ be a general point on the perpendicular bisector of $PQ$. $$\begin{aligned} PT &= QT\\ \sqrt{(x - (-13))^2 + (y - 11)^2} &= \sqrt{(x - 1)^2 + (y - 9)^2}\\ (x + 13)^2 + (y - 11)^2 &= (x - 1)^2 + (y - 9)^2\\ x^2 + 26x + 169 + y^2 - 22y + 121 &= x^2 - 2x + 1 + y^2 - 18y + 81\\ 4y &= 28x + 208\\ y &= 7x + 52 \end{aligned}$$

Or

Perpendicular bisector passes through the midpoint of chord.

Midpoint of $PQ$ $$\begin{aligned} &= \left(\frac{-13 + 1}{2},\ \frac{11 + 9}{2}\right)\\ &= (-6,\ 10)\\ m_{\perp PQ} &= 7 \end{aligned}$$ Equation of perpendicular bisector of $QR$ is $$\begin{aligned} y - 10 &= 7(x - (-6))\\ y &= 7x + 52 \end{aligned}$$

Or

Perpendicular bisector passes through the centre.

Using centre $(-7,\ 3)$ and $m_{\perp PQ} = 7$

Equation of perpendicular bisector of $QR$ is $$\begin{aligned} y - 3 &= 7(x - (-7))\\ y &= 7x + 52 \end{aligned}$$

(d)

The furthest two points on a circle is the diameter of the circle.

$\therefore$ centre $(-7, 3)$ is midpoint of diameter $QS$, $$\begin{array}{c|c} \dfrac{x_Q + x_S}{2} = -7 & \dfrac{y_Q + y_S}{2} = 3\\[6pt] 1 + x_S = -14 & 9 + y_S = 6\\[2pt] x_S = -15 & y_S = -3 \end{array}$$ $$\therefore S(-15, -3)$$ Area of quadrilateral $PQRS$ $$\begin{aligned} &= \frac{1}{2}\left| \begin{matrix} P & Q & R & S & P\\ -13 & 1 & -1 & -15 & -13\\ 11 & 9 & -5 & -3 & 11 \end{matrix} \right|\\ &= \frac{1}{2}\left|(-117 - 5 + 3 - 165) - (11 - 9 + 75 + 39)\right|\\ &= 200\text{ units}^2 \end{aligned}$$