In the previous section, we saw the basic details about conic sections. In this section, we will see some advanced details about circles.
A circle can be defined in 5 steps:
1. In fig.11.11(a) below, a curve is drawn in red color.
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A point C is marked inside the curve.
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| Fig.11.11 |
2. Mark a few convenient points P1, P2, P3, P4, P5, . . . on the curve.
• In the fig.(a), three points P1, P2 and P3 are marked.
3. Measure the distances CP1, CP2, CP3, CP4, CP5, . . .
• If all those distances are the same, then the curve is a circle.
4. The point C is called the centre of the circle.
5. The distance from C to any point on the circle is called the radius of the circle.
• It is denoted using the letter r.
Our next task is to derive the equation of a circle. It can be done in 4 steps:
1. In fig.11.12(b) above, the coordinates of C are (h,k)
2. Mark any convenient point P on the circle. Let the coordinates of P be (x,y)
3. Then the distance CP can be calculated using the distance formula. We get:
CP = $\sqrt{(x-h)^2 + (y-k)^2}$
4. But CP = r. So we can write:
$r = \sqrt{(x-h)^2 + (y-k)^2}$
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Squaring both sides, we get:
$r^2 = (x-h)^2 + (y-k)^2$
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This can be rearranged as:
$(x-h)^2 + (y-k)^2 = r^2$
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This is the equation of a circle.
The equation of a circle can be written in expanded form also. It's details can be written in 5 steps:
1. Opening the brackets in the above equation, we get:
r2 = x2 - 2hx + h2 + y2 - 2ky + k2
2. This can be rearranged as:
x2 - 2hx + h2 + y2 - 2ky + k2 - r2 = 0
3. Writing similar terms together, we get:
x2 + y2 - 2hx - 2ky + h2 + k2 - r2 = 0
4. We can describe the terms as follows:
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First term is a term in x2.
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Second term is a term in y2.
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Third term is a term in x.
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Fourth term is a term in y.
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The remaining terms are constant terms.
5. So we can write the equation as:
[x2] + [y2] + [-2hx] + [-2ky] + [h2 + k2 - r2] = 0
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Based on this we can write:
(i) The coefficient of x2 term is equal to 1.
(ii) The coefficient of y2 term is equal to 1.
(iii) The coefficient of x term is equal to -2h.
♦ So by dividing this coefficient by -2, we will get h.
♦ Thus we get the x-coordinate of the centre C.
(iv) The coefficient of y term is equal to -2k.
♦ So by dividing this coefficient by -2, we will get k.
♦ Thus we get the y-coordinate of the centre C.
(v) The constant term is h2 + k2 - r2.
♦ Once we know k, h and the constant term, we can easily calculate the radius r.
Now we will see some solved examples:
Solved example 11.1
Derive the equation of a circle whose center is at O
Solution:
1. In fig.11.11(c) above, the center of the circle is at O
2. Let P(x,y) be any point on the circle.
3. Using the distance formula, we can write:
OP = $\sqrt{(x-0)^2 + (y-0)^2}~=~\sqrt{x^2 + y^2}$
4. But OP = r. So we can write:
$r = \sqrt{x^2 + y^2}$
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Squaring both sides, we get:
$r^2 = x^2 + y^2$
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This can be rearranged as:
$x^2 + y^2 = r^2$
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This is the equation of a circle with radius r and center O.
Easier method:
1. The general equation of a circle with radius r and center at (h,k) is:
$(x-h)^2 + (y-k)^2 = r^2$
2. If the center is at O, both h and k are zero.
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Substituting these in the general equation, we get:
$(x-0)^2 + (y-0)^2 = r^2$
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This is same as: $x^2 + y^2 = r^2$
Solved example 11.2
Find the equation of the circle with center at (3/2, 7/5) and radius 5 units.
Solution:
1. The general equation of a circle with radius r and center at (h,k) is:
$(x-h)^2 + (y-k)^2 = r^2$
2. The center is given as (3/2, 7/5).
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Substituting these in the general equation, we get:
$\begin{array}{ll}
{}&{\left(x-\frac{3}{2}\right)^2 + \left(y-\frac{7}{5}\right)^2}
& {~=~}& {6^2}
&{}&{}&{} \\
{\Rightarrow}&{x^2 - 3x + \frac{9}{4}~+~y^2 - \frac{14y}{5} + \frac{49}{25}}
& {~=~}& {36}
&{}&{}&{} \\
{\Rightarrow}&{100 \times \left[x^2 - 3x + \frac{9}{4}~+~y^2 - \frac{14y}{5} + \frac{49}{25} \right]}
& {~=~36 \times 100}& {}
&{\color{green}{\text{LCM of 4, 5, 25 is 100}}}&{}&{} \\
{\Rightarrow}&{100 x^2 - 300 x + 25 × 9 + 100 y^2 - 20 × 14 × y + 4 × 49}
& {~=~3600}& {}
&{}&{}&{} \\
{\Rightarrow}&{100 x^2 + 100 y^2 - 300 x - 280 y + 225 + 196}
& {~=~3600}& {}
&{}&{}&{} \\
{\Rightarrow}&{100 x^2 + 100 y^2 - 300 x - 280 y - 3179}
& {~=~0}& {}
&{}&{}&{} \\
\end{array}$
3. The actual plot is shown in fig.11.12 below:
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| Fig.11.12 |
Solved example 11.3
Find the center and radius of the circle x2 + y2 + 8x +10y - 8 = 0
Solution:
1. The general equation of a circle is:
[x2] + [y2] + [-2hx] + [-2ky] + [h2 + k2 - r2] = 0
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The given equation can be written in the general form:
[x2] + [y2] + [8x] + [10y] + [-8] = 0
2. So we can write:
(i) coefficient of x = -2h = 8. So h = -4
(ii) coefficient of y = -2k = 10. So k = -5
(iii) constant term = h2 + k2 - r2 = -8
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Substituting the values of h and k, we get:
$\begin{array}{ll}
{}&{(-4)^2 + (-5)^2 - r^2}
& {~=~}& {-8}
&{}&{}&{} \\
{\Rightarrow}&{16 + 25 - r^2}
& {~=~}& {-8}
&{}&{}&{} \\
{\Rightarrow}&{41 - r^2}
& {~=~}& {-8}
&{}&{}&{} \\
{\Rightarrow}&{r^2}
& {~=~}& {49}
&{}&{}&{} \\
{\Rightarrow}&{r}
& {~=~}& {\sqrt{49}}
&{}&{}&{} \\
{\Rightarrow}&{r}
& {~=~}& {\pm 7}
&{}&{}&{} \\
\end{array}$
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Radius cannot be -ve. We must take the +ve value.
3. We can write:
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Center (h,k) of the circle is (-4, -5)
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Radius r of the circle is 7 units.
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Once we get the center and radius, we can use the simple form also:
$\begin{array}{ll}
{}&{(x-h)^2 + (y-k)^2}
& {~=~}& {r^2}
&{}&{}&{} \\
{\Rightarrow}&{(x~-~-4)^2 + (y~-~-5)^2}
& {~=~}& {7^2}
&{}&{}&{} \\
{\Rightarrow}&{(x+4)^2 + (y+5)^2}
& {~=~}& {49}
&{}&{}&{} \\
\end{array}$
4. The actual plot is shown below:
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| Fig.11.13 |
Solved example 11.4
Find the equation of the circle which passes through the points (2,-2) and (3,4) and whose center lies on the line x+y=2.
Solution:
1.The general equation of a circle is: (x-h)2 + (y-k)2 = r2
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Where (h,k) is the center and r is the radius.
2. In our present case, the circle passes through (2,-2). Substituting these values in the general equation, we get:
(2-h)2 + (-2-k)2 = r2
⇒ (2-h)2 + [-1(2+k)]2 = r2
⇒ (2-h)2 + (2+k)2 = r2
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Expanding this and rearranging, we get:
8 – 4h + 4k +h2 + k2 - r2 = 0
3. In our present case, the circle passes through (3,4) also. Substituting these values in the general equation, we get:
(3-h)2 + (4-k)2 = r2
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Expanding this and rearranging, we get:
25 – 6h - 8k + h2 + k2 - r2 = 0
4. So now we have two equations:
(i) From (2), we have: 8 – 4h + 4k +h2 + k2 - r2 = 0
(ii) From (3), we have: 25 – 6h - 8k + h2 + k2 - r2 = 0
5. Subtracting 4(ii) from 4(i), we get: -17 + 2h +12k = 0
6. Given that, the center lies on the line x+y=2.
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We assumed that, the center is (h,k). Substituting these in the equation of the line, we get: h+k=2
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So h = 2-k
7. Substituting this value of h in (5), we get:
-17 + 2(2-k) + 12k = 0
⇒ -17 + 4 – 2k + 12k = 0
⇒ -13 + 10k = 0
⇒ k = 1.3
8. Substituting this value of k in (6), we get: h = 2-1.3 = 0.7
9. Now we have the center of the circle (h,k) = (0.7, 1.3)
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(2,-2) is a point on the circle.
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So using the distance formula, we can write:
r2 = [(2 - 0.7)2 + (-2 - 1.3)2] = [1.32 + 3.32] = 12.58
10. Now we can write the equation of the circle:
(x - 0.7)2 + (y - 1.3)2 = 12.58
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The actual plot is shown below:
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| Fig.11.14 |
The link below gives a few more solved examples:
In the next section, we will see parabola.
