Sec 4 A Math: Coordinate Geometry of Circles, practice questions & worked solutions

(a)

$$\begin{aligned} \text{Mid-point of } AB &= \left(\frac{-3+7}{2},\,-6\right)=(2,-6) \end{aligned}$$ The $x$-coordinate of $G=2$. $$\begin{aligned} \text{Gradient of tangent at } Q &= -\frac{1}{5}\\ \text{Gradient of normal at } Q &= 5 \end{aligned}$$ Equation of normal at $Q$: $$\begin{aligned} y-(-12) &= 5(x-1)\\ y &= 5x-5-12\\ y &= 5x-17 \end{aligned}$$ The normal at $Q$ passes through the center of the circle.

When $x=2$, $y=5(2)-17=-7$. Therefore $G=(2,-7)$.

Alternative solution: $$\begin{aligned} \text{Mid-point of }AB &= \left(\frac{-3+7}{2},-6\right)=(2,-6)\\ \text{The }x\text{-coordinate of }G &= 2\\ \text{Mid-point of }AQ &= \left(\frac{-3+1}{2},\frac{-6-12}{2}\right)=(-1,-9)\\ \text{Gradient of }AQ &= \frac{-6-(-12)}{-3-1}=-\frac{3}{2}\\ \text{Equation of perpendicular bisector of }AQ: y+9 &= \frac{2}{3}(x+1)\\ y &= \frac{2}{3}x-8\frac{1}{3}\\ x=2, y &= \frac{4}{3}-8\frac{1}{3}=-7\\ G &= (2,-7) \end{aligned}$$

(b)

$$\begin{aligned} \text{Distance of } GA &= \sqrt{(2-(-3))^2+(-7-(-6))^2}\\ &= \sqrt{26} \end{aligned}$$ Equation of circle: $$\begin{aligned} (x-2)^2+(y+7)^2 &= (\sqrt{26})^2\\ x^2-4x+4+y^2+14y+49 &= 26\\ x^2-4x+y^2+14y+27 &= 0\\ g=-2,\ f=7,\ &c=27 \end{aligned}$$

(a)

$$\begin{aligned} \text{Midpoint of } PQ &= \left(\frac{-6 + 2}{2},\ \frac{-12 + (-20)}{2}\right)\\ &= (-2, -16)\\ \text{Gradient } PQ &= \frac{-12 + 20}{-6 - 2} = -1 \end{aligned}$$ Equation of perpendicular bisector of $PQ$ is $$\begin{aligned} y + 16 &= 1(x + 2)\\ y &= x - 14 \text{----- (1)}\\ 7y + 4x + 87 &= 0 \text{----- (2)} \end{aligned}$$ Substitute (1) into (2): $$\begin{aligned} 7x - 98 + 4x + 87 &= 0\\ x = 1 \Rightarrow y &= -13\\ A &= (1, -13) \end{aligned}$$ $$\begin{aligned} \text{Radius} &= \sqrt{(1 - 2)^2 + \left(-13 - (-20)\right)^2}\\ &= \sqrt{1^2 + 7^2}\\ &= \sqrt{50}\\ &= 5\sqrt{2}\ \text{units} \end{aligned}$$

(b)

$$(x - 1)^2 + (y + 13)^2 = 50$$

(c)

New centre $= (-13, 1)$

Equation of circle $C_2$ is $$(x + 13)^2 + (y - 1)^2 = 50$$

(a)

Centre of $C_1 = (2, y)$

Grad of normal $= \dfrac{3}{4}$ $$\begin{aligned} \frac{y - 6}{2 - 6} &= \frac{3}{4}\\ y &= \frac{3}{4}(-4) + 6\\ &= 3 \end{aligned}$$ $\therefore$ Centre of $C_1 = (2, 3)$

Radius $= 5$ units

Equation of $C_1$: $(x - 2)^2 + (y - 3)^2 = 25$

(b)

Let the centre $C_2 = (x, y)$ $$\left(\frac{x + 2}{2},\ \frac{y + 3}{2}\right) = (6, 6)$$ $$x = 10,\ y = 9$$ Equation of $C_2$: $(x - 10)^2 + (y - 9)^2 = 25$

(a)

Perpendicular bisector of chord passes through the centre of circle.

Hence, since $AB$ is a horizontal chord, $P$ is vertically above/below the midpoint of $AB$. $$\therefore\; x_P=\frac{1+9}{2}=5$$

(b)

$$\begin{aligned} 4y+3(5) &= 31\\ 4y &= 16\\ y &= 4 \end{aligned}$$ $P$ is $(5,\,4)$. $$\begin{aligned} r &= \sqrt{(9-5)^2+(7-4)^2}\\ &= \sqrt{25} = 5 \end{aligned}$$ Equation is circle is $(x-5)^2+(y-4)^2=25$.

(c)

Gradient of normal to $A$ is $-\dfrac{3}{4}$, hence gradient of tangent at $A$ is $\dfrac{4}{3}$.

Tangent: $y=\dfrac{4}{3}x+c$. $$\begin{aligned} \text{At } A(1,7): 7 &= \frac{4}{3}+c\\ c &= \frac{17}{3} \end{aligned}$$ $$y=\frac{4}{3}x+\frac{17}{3}$$ $D$ is $\left(0,\,\dfrac{17}{3}\right)$.

(d)

$$\begin{aligned} &T(1-8,\,7+6)\\ &T(-7,\,13) \end{aligned}$$

(a)

$y$-coordinate of centre $=\dfrac{-1+9}{2}=4$.

(b)

Let the $x$-coordinate of the centre be $a$, $$\begin{aligned} (x-a)^2+(y-4)^2 &= 25 \end{aligned}$$ From the line, $y=\dfrac{4x+9}{3}$: $$\begin{aligned} (x-a)^2+\left(\frac{4x+9}{3}-4\right)^2 &= 25\\ (x-a)^2+\left(\frac{4}{3}x-1\right)^2 &= 25\\ x^2-2ax+a^2+\frac{16}{9}x^2-\frac{8}{3}x+1 &= 25\\ \frac{25}{9}x^2+\left(-2a-\frac{8}{3}\right)x+a^2-24 &= 0 \end{aligned}$$ Since the line is a tangent, $b^2-4ac=0$: $$\begin{aligned} \left(-2a-\frac{8}{3}\right)^2-4\!\left(\frac{25}{9}\right)(a^2-24) &= 0\\ 4a^2+\frac{32}{3}a+\frac{64}{9}-\frac{100}{9}a^2+\frac{2400}{9} &= 0\\ 36a^2+96a+64-100a^2+2400 &= 0\\ -64a^2+96a+2464 &= 0\\ 2a^2-3a-77 &= 0\\ (2a+11)(a-7) &= 0\\ a=7 \;&\text{or}\; -5.5\ (\text{rej}) \end{aligned}$$

(c)

$$\begin{aligned} \text{Distance between point and centre} &= \sqrt{(7-3)^2+(4-0)^2}\\ &= 5.66\ \text{unit} \end{aligned}$$ Since the distance is longer than the radius, the point lies outside.

(a)

$$(x - 2)^2 + (y + 3)^2 = 20 \text{OR} x^2 + y^2 - 4x + 6y - 7 = 0$$

(b)

$$y = -3 + \sqrt{20} \& y = -3 - \sqrt{20}$$

(c)

At $y$-axis, $x = 0$, $$\begin{aligned} \therefore y &= 1 \text{ or } -7\\ ST &= 1 + 7 = 8\ \text{units} \end{aligned}$$

(d)

Perpendicular bisector of $ST$ will pass through the centre of the circle. since $ST$ is vertical. $$\begin{aligned} y\text{-coord of } B &= y\text{ coord of } A\\ &= \frac{1 - 7}{2} = -3 \end{aligned}$$

(e)

using $(0, -7)$ & $B(x, -3)$ $$\begin{aligned} x^2 + (-7 + 3)^2 &= 52\\ x &= 6 \ \text{or}\ -6 \ (\text{rej since } x > 0) \end{aligned}$$ $x$-coord of $B = 6$

(a)

$$\begin{aligned} \text{Gradient of } DE &= \frac{-3 + 7}{-4 - 4} = -\frac{1}{2}\\ \text{Gradient of normal to } DE &= 2\\ \text{Midpoint of } DE &= \left(\frac{-4 + 4}{2},\ \frac{-3 - 7}{2}\right) = (0, -5) \end{aligned}$$ Equation of normal: $y = 2x - 5$ $$\begin{aligned} 4(2x - 5) &= 3x - 15\\ 5x &= 5\\ x &= 1\\ y &= 2 - 5 = -3 \end{aligned}$$ $$\begin{aligned} \text{Centre} &= (1, -3)\\ \text{Radius} &= \sqrt{(1 + 4)^2 + (-3 + 3)^2} = 5 \text{ units} \end{aligned}$$ Equation of circle: $$\begin{aligned} (x - 1)^2 + (y + 3)^2 &= 5^2\\ (x - 1)^2 + (y + 3)^2 &= 25 \end{aligned}$$

(b)

Distance between $(-1, 1)$ and $(1, -3)$ $$\begin{aligned} &= \sqrt{(1 + 1)^2 + (-3 - 1)^2}\\ &= 4.47 < 5 \end{aligned}$$ Since the distance between $(-1, 1)$ and $(1, -3)$ is lesser than the radius of the circle, the point $(-1, 1)$ lies inside the circle.

(c)

Solving simultaneously $4y = 3x - 15$ and $y = x - 3$: $$\begin{aligned} 4(x - 3) &= 3x - 15\\ x &= -3\\ y &= -6 \end{aligned}$$ Since the point of intersection of $y = x - 3$ and the normal to the circle is not $(1, -3)$ but $(-3, -6)$, $AB$ is not a possible diameter of the circle.

Alternatively,

As $(1, -3)$ is the centre of the circle, sub. $x = 1$ into the line $y = x - 3$. $$y = 1 - 3 = -2$$ Since $y \neq -3$, the line $y = x - 3$ does not pass through the centre $(1, -3)$.

Therefore, the line segment $PQ$ is not a possible diameter of the circle.

(a)

$$\begin{aligned} m_t &= \frac{3 - 0}{-3 - 0}\\ &= -1 \end{aligned}$$ Gradient of normal $= 1$

Equation of normal is $y - 3 = 1(x + 3)$ $$y = x + 6$$

(b)

$$\begin{aligned} \text{Equation of normal is } y &= -\frac{2}{5}x - 1 \text{--- (1)}\\ y &= x + 6 \text{--- (2)} \end{aligned}$$ Sub (1) into (2) $$\begin{aligned} -\frac{2}{5}x - 1 &= x + 6\\ \frac{7}{5}x &= -7\\ x &= -5 \end{aligned}$$ $$\text{Centre} = (-5, 1)$$ $$\begin{aligned} r^2 &= (-3 + 5)^2 + (3 - 1)^2\\ &= 8 \end{aligned}$$ Therefore equation of circle is $(x + 5)^2 + (y - 1)^2 = 8$.

(c)

$$\left(-5 - 2\sqrt{2},\ 1\right)$$

(a)

$$\begin{aligned} x^2 + y^2 - 8x - 4y + 15 &= 0\\ x^2 - 8x + 4^2 + y^2 - 4y + 2^2 &= 4^2 + 2^2 - 15\\ (x - 4)^2 + (y - 2)^2 &= 5 \end{aligned}$$ Radius is $\sqrt{5}$ units

Centre $(4, 2)$

(b)

$$\begin{aligned} m_{AB} &= \frac{-2 - 2}{5 - 9}\\ &= 1\\ \text{Midpoint of } AB &= \left(\frac{5 + 9}{2},\ \frac{-2 + 2}{2}\right)\\ &= (7, 0) \end{aligned}$$ Equation of perpendicular bisector is $$\begin{aligned} y - 0 &= -1(x - 7)\\ y &= -x + 7 \end{aligned}$$ Substituting equation of line into equation of circle: $$\begin{aligned} (x - 4)^2 + (-x + 7 - 2)^2 &= 5\\ x^2 - 8x + 16 + x^2 - 10x + 25 &= 5\\ 2x^2 - 18x + 36 &= 0\\ x^2 - 9x + 18 &= 0\\ (x - 3)(x - 6) &= 0\\ x = 3 &\text{or} x = 6\\ y = 4 &\phantom{\text{or} } y = 1 \end{aligned}$$ The coordinates are $(3, 4)$ and $(6, 1)$.

(c)

$$\begin{aligned} \text{Area of } OCD &= \frac{1}{2}\left| \begin{matrix} 0 & 6 & 3 & 0\\ 0 & 1 & 4 & 0 \end{matrix} \right|\\ &= \frac{1}{2}\left[(0 + 24 + 0) - (0 + 3 + 0)\right]\\ &= 10.5\\ \text{Length of } CD &= \sqrt{(3 - 6)^2 + (4 - 1)^2}\\ &= \sqrt{18}\\ &= 3\sqrt{2}\\ \text{shortest dist} &= \frac{10.5}{\frac{1}{2} \times 3\sqrt{2}}\\ &= \frac{7}{\sqrt{2}}\\ &= \frac{7}{2}\sqrt{2} \end{aligned}$$

Alternatively,

The perpendicular line from $O$ to $CD$ has gradient $1$, since the gradient of $CD$ is $-1$.

Thus the perpendicular line has equation $y = x$, since the $y$-intercept is $0$.

Intersecting the 2 lines $y = x$ and $y = -x + 7$, we have $$\begin{aligned} x &= -x + 7\\ 2x &= 7\\ x &= \frac{7}{2}\\ y &= \frac{7}{2} \end{aligned}$$ The point on $CD$ shortest distance from $O$ is $\left(\dfrac{7}{2},\ \dfrac{7}{2}\right)$.

Thus, the distance is $$\begin{aligned} \sqrt{\left(\frac{7}{2} - 0\right)^2 + \left(\frac{7}{2} - 0\right)^2} &= \sqrt{2\left(\frac{7}{2}\right)^2}\\ &= \frac{7}{2}\sqrt{2} \end{aligned}$$