Conic Sections
Exercise 10.1 • Circles
Standard Equation of a Circle:
$(x – h)^2 + (y – k)^2 = r^2$
Where $(h, k)$ is the centre and $r$ is the radius.
$(x – h)^2 + (y – k)^2 = r^2$
Where $(h, k)$ is the centre and $r$ is the radius.
Q1-Q5
Find the equation of the circle with the given parameters:
1. Centre (0, 2) and radius 2
Substitute
$(x – 0)^2 + (y – 2)^2 = 2^2$
Expand
$x^2 + y^2 – 4y + 4 = 4 \Rightarrow x^2 + y^2 – 4y = 0$
2. Centre (-2, 3) and radius 4
Substitute
$(x – (-2))^2 + (y – 3)^2 = 4^2$
Expand
$(x + 2)^2 + (y – 3)^2 = 16 \Rightarrow x^2 + 4x + 4 + y^2 – 6y + 9 = 16$
Simplify
$x^2 + y^2 + 4x – 6y – 3 = 0$
3. Centre (1/2, 1/4) and radius 1/12
Substitute
$(x – \frac{1}{2})^2 + (y – \frac{1}{4})^2 = (\frac{1}{12})^2$
Expand
$x^2 – x + \frac{1}{4} + y^2 – \frac{y}{2} + \frac{1}{16} = \frac{1}{144}$
Multiply
Multiply by 144: $144x^2 – 144x + 36 + 144y^2 – 72y + 9 = 1$
Result
$144x^2 + 144y^2 – 144x – 72y + 44 = 0$ (Divide by 4: $36x^2+36y^2-36x-18y+11=0$)
4. Centre (1, 1) and radius $\sqrt{2}$
Equation
$(x-1)^2 + (y-1)^2 = (\sqrt{2})^2 = 2$
Expand
$x^2 – 2x + 1 + y^2 – 2y + 1 = 2 \Rightarrow x^2 + y^2 – 2x – 2y = 0$
5. Centre (-a, -b) and radius $\sqrt{a^2-b^2}$
Substitute
$(x – (-a))^2 + (y – (-b))^2 = (\sqrt{a^2-b^2})^2$
Expand
$(x+a)^2 + (y+b)^2 = a^2 – b^2 \Rightarrow x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 – b^2$
Simplify
$x^2 + y^2 + 2ax + 2by + 2b^2 = 0$
Q6-Q9
Find the centre and radius of the circles:
6. $(x + 5)^2 + (y – 3)^2 = 36$
Compare
$(x – h)^2 + (y – k)^2 = r^2$
Result
$h = -5, k = 3, r^2 = 36 \Rightarrow r = 6$.
Centre (-5, 3), Radius 6
Centre (-5, 3), Radius 6
7. $x^2 + y^2 – 4x – 8y – 45 = 0$
General Eq
$x^2 + y^2 + 2gx + 2fy + c = 0$. Centre $(-g, -f)$, $r = \sqrt{g^2+f^2-c}$
Values
$2g = -4 \Rightarrow g = -2$. $2f = -8 \Rightarrow f = -4$. $c = -45$.
Calc
Centre: $(2, 4)$. Radius: $\sqrt{(-2)^2 + (-4)^2 – (-45)} = \sqrt{4+16+45} = \sqrt{65}$.
8. $x^2 + y^2 – 8x + 10y – 12 = 0$
Values
$2g = -8 \Rightarrow g = -4$. $2f = 10 \Rightarrow f = 5$. $c = -12$.
Calc
Centre: $(4, -5)$. Radius: $\sqrt{16 + 25 – (-12)} = \sqrt{53}$.
9. $2x^2 + 2y^2 – x = 0$
Normalize
Divide by 2: $x^2 + y^2 – \frac{1}{2}x = 0$.
Values
$2g = -1/2 \Rightarrow g = -1/4$. $f = 0$. $c = 0$.
Result
Centre: $(1/4, 0)$. Radius: $\sqrt{(1/4)^2 + 0 – 0} = 1/4$.
Q10
10. Find equation: Passes through (4,1) and (6,5), centre on 4x + y = 16.
Let
Centre be $(h, k)$ and radius $r$. Equation: $(x-h)^2 + (y-k)^2 = r^2$.
Point 1
$(4-h)^2 + (1-k)^2 = r^2 \dots (1)$
Point 2
$(6-h)^2 + (5-k)^2 = r^2 \dots (2)$
Equate
Expand and equate (1) & (2): $-8h -2k + 17 = -12h -10k + 61 \Rightarrow 4h + 8k = 44 \Rightarrow h + 2k = 11$.
Line
Centre on line: $4h + k = 16$.
Solve
Solving $h+2k=11$ and $4h+k=16$ gives $h=3, k=4$. Centre $(3,4)$.
Radius
$r^2 = (4-3)^2 + (1-4)^2 = 1 + 9 = 10$.
$(x-3)^2 + (y-4)^2 = 10 \Rightarrow x^2 + y^2 – 6x – 8y + 15 = 0$
Q11
11. Find equation: Passes through (2,3) and (-1,1), centre on x – 3y – 11 = 0.
Equate
$(2-h)^2 + (3-k)^2 = (-1-h)^2 + (1-k)^2$
Simplify
$-4h -6k + 13 = 2h -2k + 2 \Rightarrow 6h + 4k = 11$.
System
1) $6h + 4k = 11$.
2) $h – 3k = 11$ (Centre lies on line).
2) $h – 3k = 11$ (Centre lies on line).
Solve
$h = 7/2, k = -5/2$.
Radius
$r^2 = (2 – 3.5)^2 + (3 – (-2.5))^2 = (-1.5)^2 + (5.5)^2 = 2.25 + 30.25 = 32.5$.
$(x – \frac{7}{2})^2 + (y + \frac{5}{2})^2 = \frac{65}{2}$
Q12
12. Radius 5, centre on x-axis, passes through (2,3).
Centre
Centre is $(h, 0)$ because it lies on x-axis.
Distance
Distance from $(h,0)$ to $(2,3)$ is 5. $(h-2)^2 + (0-3)^2 = 25$.
Solve
$(h-2)^2 + 9 = 25 \Rightarrow (h-2)^2 = 16 \Rightarrow h-2 = \pm 4$.
Values
$h = 6$ or $h = -2$.
$(x-6)^2 + y^2 = 25$ OR $(x+2)^2 + y^2 = 25$
Q13
13. Passes through (0,0) and intercepts a and b on axes.
Points
Circle passes through $O(0,0)$, $A(a,0)$, and $B(0,b)$.
Geometry
Since $\angle AOB = 90^\circ$, AB is the diameter.
Centre
Midpoint of AB: $(\frac{a+0}{2}, \frac{0+b}{2}) = (\frac{a}{2}, \frac{b}{2})$.
Radius
Distance from Centre to Origin: $\sqrt{(a/2)^2 + (b/2)^2}$. $r^2 = \frac{a^2+b^2}{4}$.
Equation
$(x – a/2)^2 + (y – b/2)^2 = \frac{a^2+b^2}{4}$.
$x^2 + y^2 – ax – by = 0$
Q14
14. Centre (2,2) and passes through (4,5).
Radius
$r^2 = (4-2)^2 + (5-2)^2 = 2^2 + 3^2 = 4 + 9 = 13$.
Equation
$(x-2)^2 + (y-2)^2 = 13$.
Expand
$x^2 – 4x + 4 + y^2 – 4y + 4 = 13$.
$x^2 + y^2 – 4x – 4y – 5 = 0$
Q15
15. Does (-2.5, 3.5) lie inside, outside or on $x^2 + y^2 = 25$?
[Image showing point position relative to a circle boundary]
Compare
Calculate distance of point from center (0,0). Compare with Radius 5.
Calc
$d^2 = (-2.5)^2 + (3.5)^2 = 6.25 + 12.25 = 18.5$.
Result
$18.5 < 25$. Distance is less than radius squared.
The point lies inside the circle.