Circle

Example

Write the standard equation of the circle given:

1. Centre $(3, 2)$ and radius 4 units.
2. Centre $(0, 4)$ and diameter 6 units.

Solution:

a. Given centre $(a, b) = (3, 2)$ and radius $r = 4$, substitute into the standard equation:

b. Given centre $(0, 4)$ and diameter $d = 6$, the radius is $r = \frac{d}{2} = 3$. Substitute:

The general equation of a circle arises from expanding the standard form and rearranging terms. Starting from equation (4.12):

Expanding the squares:

Rearranging all terms to one side gives:

To simplify notation, define:

• $g = -a$
• $f = -b$
• $c = a^2 + b^2 - r^2$

Thus, the general equation of a circle can be expressed as:

Important observations

• The coefficients of $x^2$ and $y^2$ must be equal and positive (both 1) for the equation to represent a circle.
• The coefficients of $x$ and $y$ terms ($2g$ and $2f$ respectively) are related to the coordinates of the centre.
• The constant term $c$ is related to the radius and the coordinates of the centre.

Finding centre and radius from general equation

Given the general form:

The centre and radius are:

Example

Find the centre and radius of the circle given by the equation:

Solution:

Compare with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$:

• $2g = -4 \Rightarrow g = -2$
• $2f = 2 \Rightarrow f = 1$
• $c = 1$

Find the centre:

Calculate the radius:

Therefore, the circle's centre is $(2, -1)$ and radius is 2 units.

Alternative method: Complete the square for both $x$ and $y$ terms:

This confirms the centre and radius found.

Example

Find the general equation of a circle with centre $(1, 3)$ and radius 2 units.

Solution:

Given centre $(a, b) = (1, 3)$, we have $g = -a = -1$ and $f = -b = -3$.

Using the formula for $c$:

Substitute values into the general form:

This is the required general equation of the circle.

• The standard equation of a circle explicitly encodes the geometric definition of a circle in analytic geometry, making it straightforward to determine if a point lies on the circle.
• The general equation is particularly useful for algebraic manipulation, curve sketching, and understanding how transformations affect the circle's position and size.
• Completing the square is a vital algebraic technique to convert the general form back to standard form, which directly reveals the centre and radius.
• Understanding the relationship between the coefficients in the general equation and the circle's geometric parameters is fundamental for problems involving tangents, chords, and intersections with other geometric shapes.
• In advanced applications, circles can be generalized to spheres in 3D, or generalized conic sections where coefficients do not satisfy the circle conditions, leading to ellipses, parabolas, or hyperbolas.

Let AB be the diameter of a circle with endpoints at coordinates $A(x_1, y_1)$ and $B(x_2, y_2)$. For any point $P(x, y)$ lying on this circle, the angle $\widehat{APB}$ subtended by the diameter at the circumference is a right angle, i.e., $90^\circ$. This is a direct consequence of the Thales' theorem, which states that any angle inscribed in a semicircle is a right angle.

Figure 1: Circle with diameter AB

In coordinate geometry, the gradients (slopes) of the line segments $AP$ and $BP$ satisfy a perpendicularity condition since $\widehat{APB} = 90^\circ$. The gradients are given by:

Since $AP \perp BP$, their gradients satisfy:

Substituting the expressions for $m_1$ and $m_2$, we have:

Cross-multiplying and rearranging leads to the equation:

Interpretation: This represents the locus of all points $P(x, y)$ that lie on the circle with diameter $AB$.

Example

Problem: Find the equation of the circle with diameter endpoints $A(2, 1)$ and $B(-2, 2)$.

Solution:

Given $A(x_1, y_1) = (2,1)$ and $B(x_2, y_2) = (-2, 2)$, substitute into equation (1):

Thus, the equation of the circle is:

Consider three distinct points $A_1(x_1, y_1)$, $A_2(x_2, y_2)$, and $A_3(x_3, y_3)$, which lie on a circle with center $C(a, b)$ and radius $r$. The standard form of the circle's equation is:

Since each point lies on the circle, substituting their coordinates yields three equations:

Subtracting pairs of these equations eliminates $r^2$, enabling the determination of $a$ and $b$, the coordinates of the center. Subsequently, substituting $a$ and $b$ back into one of the equations yields the radius $r$.

Figure 2: Circle passing through points $A_1$, $A_2$, and $A_3$

Alternative approach: Use the general form of a circle's equation:

Substituting the three points produces a system of linear equations in terms of $g$, $f$, and $c$:

Solving this system gives the parameters $g, f, c$. The center and radius can then be found using:

Example

Problem: Find the equation of the circle passing through the points $A(3, 1)$, $B(2, 6)$, and $C(3, 2)$.

Solution:

Using the general form $x^2 + y^2 + g x + f y + c = 0$, substitute each point:

Simplify:

Subtract the first from the third and the first from the second to eliminate $c$:

Substitute $f = -3$ into second equation:

Use the first equation to find $c$:

Final equation:

Center and radius:

Thus, the circle's center is $(-7.5, 1.5)$ and radius $\approx 10.52$.

A tangent line to a circle is a straight line that touches the circle at exactly one point. This unique point is called the point of tangency. A crucial geometric property of the tangent is that it is always perpendicular to the radius drawn to the point of tangency.

General case: deriving the tangent equation

Consider a circle with center $C(-g, -f)$ and a point of tangency $P(x_0, y_0)$ on its circumference. Let the point $Q(x, y)$ be any other point on the tangent line. Since the radius $\overline{CP}$ is perpendicular to the tangent at $P$, the slopes of $\overline{CP}$ and the tangent line are negative reciprocals of each other.

The slope of the radius $\overline{CP}$ is:

Therefore, the slope of the tangent line is the negative reciprocal of the slope of CP:

Using the point-slope form of a line, the equation of the tangent at point $(x_0, y_0)$ is:

Alternatively, using the geometric condition that the line is perpendicular to the radius and passes through $(x_0, y_0)$, the equation of the tangent line to the circle is also given by:

Example 1: tangent line to a given circle

Find the equation of the tangent to the circle

at the point $P(1, -4)$.

Solution

1. Compare the given circle with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

2. We find:

   $$2g = -2 \Rightarrow g = -1, \quad 2f = 4 \Rightarrow f = 2$$

   So the center is $C(1, -2)$.

3. The slope of radius $CP$, from $C(1, -2)$ to $P(1, -4)$, is:

   $$m_{CP} = \frac{-4 + 2}{1 - 1} = \frac{-2}{0}$$

   This is undefined, indicating a vertical radius.

4. Since the tangent is perpendicular to the radius, and the radius is vertical, the tangent must be a horizontal line.

5. Thus, the equation of the tangent is simply:

   $$y = -4$$

Example 2: proving tangent condition

Show that the line $y = -mx + c$ is tangent to the circle $x^2 + y^2 = a^2$ if and only if:

Solution

1. Substitute $y = -mx + c$ into the circle equation:

   $$x^2 + (-mx + c)^2 = a^2$$ $$x^2 + m^2x^2 - 2mcx + c^2 = a^2$$ $$(1 + m^2)x^2 - 2mcx + (c^2 - a^2) = 0$$

2. This is a quadratic in $x$. For the line to be a tangent, it must intersect the circle at exactly one point. Therefore, the discriminant must be zero:

   $$\Delta = (-2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0$$ $$4m^2c^2 - 4(1 + m^2)(c^2 - a^2) = 0$$ $$m^2c^2 = (1 + m^2)(c^2 - a^2)$$

3. Expanding and simplifying gives:

   $$m^2c^2 = c^2 + m^2c^2 - a^2 - a^2m^2$$ $$c^2 = a^2(1 + m^2)$$ $$\Rightarrow c = \pm a\sqrt{1 + m^2}$$

Example 3: tangents parallel to a given line

Find the equations of the tangents to the circle

that are parallel to the line $4x + 3y + 5 = 0$.

Solution

1. Any line parallel to $4x + 3y + 5 = 0$ has the form:

   $$4x + 3y + c = 0$$

2. Rewrite the circle in standard form:

   $$(x - 3)^2 + (y + 2)^2 = 25$$

   So, the center is $(3, -2)$, and the radius is 5.

3. Use the perpendicular distance formula:

   $$\frac{|4(3) + 3(-2) + c|}{\sqrt{4^2 + 3^2}} = 5$$ $$\frac{|12 - 6 + c|}{5} = 5 \Rightarrow |c + 6| = 25$$ $$c + 6 = \pm 25 \Rightarrow c = 19 \text{ or } c = -31$$

4. Therefore, the equations of the tangents are:

   $$4x + 3y + 19 = 0 \quad \text{and} \quad 4x + 3y - 31 = 0$$

The normal to a circle is defined as the straight line that is drawn perpendicular (i.e., at 90°) to the tangent at the point where the tangent touches the circle.

Geometrically, if a point $P(x_0, y_0)$ lies on a circle, and a tangent is drawn to the circle at point $P$, then the normal is the line that intersects the circle at $P$ and is perpendicular to the tangent at that point. This relationship is illustrated in the diagram below.

Figure: Normal and Tangent to a Circle at Point P

Let the equation of the circle be given in standard form:

From this equation, the centre of the circle is located at:

Let $P(x_0, y_0)$ be a point on the circle (i.e., a point of tangency), and suppose the line normal to the circle passes through this point. The key geometric insight is:

• The normal is collinear with the line joining the centre $C(-g, -f)$ to the point of tangency $P(x_0, y_0)$.
• Hence, the slope of the normal is equal to the slope of the line segment $CP$.

The slope $m$ of line $CP$ is computed as:

Since the normal passes through the point $P(x_0, y_0)$ and has slope $m$, we use the point-slope form of a linear equation to write:

This is the general equation of the normal to the circle at the point $(x_0, y_0)$.

Example 4.26

Problem: Find the equation of the normal to the circle

at the point $P(-1, -5)$.

Solution

First, we compare the given equation with the general form:

From the comparison, we extract the values:

• $2g = -6 \Rightarrow g = -3$
• $2f = 4 \Rightarrow f = 2$

So, the centre of the circle is:

Given point of tangency: $P(-1, -5)$

The slope $m$ of the normal (i.e., of line $CP$) is:

The point-slope form of the line passing through $P(-1, -5)$ with slope $m = -\frac{3}{2}$ is:

Simplifying:

Multiplying through by 2 to eliminate the fraction:

Answer: The equation of the normal is:

There are three main cases in which two circles can intersect:

1. External tangency – The circles touch each other externally.
2. Internal tangency – The circles touch each other internally.
3. Intersection at two distinct points – The circles intersect at two separate points.

i. External point of intersection

Two circles touch each other externally if and only if the distance between their centres is equal to the sum of their radii.

Let the equations of two circles be:

• Circle 1: $(x - a_1)^2 + (y - b_1)^2 = r_1^2$
• Circle 2: $(x - a_2)^2 + (y - b_2)^2 = r_2^2$

Then the distance between the centres is:

Condition for external intersection:

Example 4.27

Show that the circles:

Solution:

From the first circle: Center = $C_1(2, 3)$, Radius = $r_1 = \sqrt{2^2 + 3^2 - 3} = \sqrt{16} = 4$

From the second circle: Center = $C_2(6, 6)$, Radius = $r_2 = \sqrt{6^2 + 6^2 - 71} = \sqrt{1} = 1$

Distance between centres:

Sum of radii: $r_1 + r_2 = 4 + 1 = 5$

Since $C_1C_2 = r_1 + r_2$, the circles intersect externally.

ii. Internal point of intersection

Two circles touch each other internally if and only if the distance between their centres is equal to the difference of their radii.

Condition for internal intersection:

Example 4.28

Show that the circles:

Solution:

First circle: Center = $C_1(-1, 0)$, Radius = $r_1 = \sqrt{1 + 0 + 8} = \sqrt{9} = 3$

Second circle: Center = $C_2(3, -3)$, Radius = $r_2 = \sqrt{9 + 9 + 46} = \sqrt{64} = 8$

Distance between centres:

Difference of radii: $|8 - 3| = 5$

Since $C_1C_2 = |r_2 - r_1|$, the circles intersect internally.

iii. Intersection at two distinct points

Two circles intersect at two distinct points if and only if the distance between their centres is less than the sum of their radii and greater than their difference:

$|r_1 - r_2| < C_1C_2 < r_1 + r_2$

The points of intersection satisfy both circle equations. The line joining these points is called the common chord.

To find the equation of the common chord:

Given:

• $C_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$
• $C_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$

Subtracting the equations:

Example

Find the equation of the common chord and the points of intersection of the circles:

Solution:

Subtracting second from the first:

So, common chord is:

Substituting into first circle:

Corresponding x-values:

If $y = 2$, $x = 3(2) - 8 = -2$

If $y = 3$, $x = 3(3) - 8 = 1$

Points of intersection:

Orthogonal circles

Two circles are said to be orthogonal if they intersect in such a way that the tangents at their points of intersection are perpendicular to each other, as shown in Figure 4.21.

Figure 4.21: Orthogonal circles intersecting at right angles

Let the equations of the two circles be:

Then the centres and radii of the two circles are:

• Centre of $C_1$: $(-g_1, -f_1)$, radius $r_1 = \sqrt{g_1^2 + f_1^2 - c_1}$
• Centre of $C_2$: $(-g_2, -f_2)$, radius $r_2 = \sqrt{g_2^2 + f_2^2 - c_2}$

Let $d$ be the distance between the centres of the two circles. Then:

The two circles are orthogonal if and only if the square of the distance between their centres equals the sum of the squares of their radii. That is:

Example

Prove that the two circles

are orthogonal.

Solution

From the given equations:

• Centre of $C_1$: $(4, 7)$, radius $r_1 = \sqrt{4^2 + 7^2 - 56} = \sqrt{65 - 56} = \sqrt{9} = 3$
• Centre of $C_2$: $(2, 4)$, radius $r_2 = \sqrt{2^2 + 4^2 - 16} = \sqrt{20 - 16} = \sqrt{4} = 2$

The distance between the centres is:

Now, calculate $r_1^2 + r_2^2$:

Since $d^2 = r_1^2 + r_2^2 = 13$, the two circles are orthogonal.

Length of a tangent from a point to a circle

Consider the general equation of a circle:

This circle has:

• Centre: $C(-g, -f)$
• Radius: $r = \sqrt{g^2 + f^2 - c}$, assuming $g^2 + f^2 - c > 0$

Let $P(x_1, y_1)$ be a point outside the circle, as shown in Figure 4.22. The tangent from point $P$ to the circle touches the circle at point $T$, and the line $CT$ is perpendicular to the tangent.

Figure 4.22: Tangent from a point to a circle

By Pythagoras' theorem in triangle $\triangle CTP$, we have:

From the geometry of the circle:

• $CT = r = \sqrt{g^2 + f^2 - c} \Rightarrow CT^2 = g^2 + f^2 - c$ (4.24)
• $CP = \sqrt{(x_1 + g)^2 + (y_1 + f)^2} \Rightarrow CP^2 = (x_1 + g)^2 + (y_1 + f)^2$ (4.25)

Substituting (4.24) and (4.25) into (4.23):

Expanding and simplifying:

Therefore, the length of the tangent is:

Example

Find the length of the tangent from point $P(2, 5)$ to the circle:

Solution

We compare with the general form:

Using the formula:

Answer: The length of the tangent is 3 units.