Higher Secondary Mathematics: Chapter 11.1 - General Equation of A Circle

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.
• A point C is marked inside the curve.

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:
$r = \sqrt{(x-h)^2 + (y-k)^2}$
• Squaring both sides, we get:
$r^2 = (x-h)^2 + (y-k)^2$
• This can be rearranged as:
$(x-h)^2 + (y-k)^2 = r^2$
• 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:
• First term is a term in x².
• Second term is a term in y².
• Third term is a term in x.
• Fourth term is a term in y.
• The remaining terms are constant terms.
5. So we can write the equation as:
[x²] + [y²] + [-2hx] + [-2ky] + [h² + k² - r²] = 0
• 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:
$r = \sqrt{x^2 + y^2}$
• Squaring both sides, we get:
$r^2 = x^2 + y^2$
• This can be rearranged as:
$x^2 + y^2 = r^2$
• 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.
• Substituting these in the general equation, we get:
$(x-0)^2 + (y-0)^2 = r^2$
• 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).
• 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:

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
• 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
• 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}$

• Radius cannot be -ve. We must take the +ve value.
3. We can write:
• Center (h,k) of the circle is (-4, -5)
• Radius r of the circle is 7 units.
• 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:

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
• 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
• 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
• 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.
• We assumed that, the center is (h,k). Substituting these in the equation of the line, we get: h+k=2
• 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)
• (2,-2) is a point on the circle.
• 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
• The actual plot is shown below: