The Equation of a Circle

A circle is defined as the set of points in a plane that are equidistant from a fixed point called the centre, $O$. Below are some basic definitions related to the geometry of a circle which you should know:

• The circumference is the distance around the perimeter of a circle. It can be calculated with the formula $2\pi r$.
• The radius, $r$, is the distance from the centre of the circle to any point on its circumference.
• The diameter, $d$, is the length of a straight line segment which passes through the centre of the circle and whose two endpoints are on the circumference of the circle. The diameter is twice the radius. ($d = 2r$)

As required by the syllabus, you need to know the equation of the circle. That is, how to define a circle in terms of $x$ and $y$ on the Cartesian plane. Let $O$ be the centre of the circle, fixed at the coordinates $(h,k)$ where $h,k \in \mathbb{R}$.

Based on our definition of the circle, all the points $(x,y)$ which are on the circle must satisfy the condition that the distance between $(x,y)$ and $O(h,k)$ is always some constant value. Recall that the radius, $r$, is this constant distance.

Recall that the distance between any two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by the formula $$|PQ| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

Since $r$ is the distance between $(x,y)$ and $O(h,k)$, $$r = \sqrt{(x-h)^2 + (y-k)^2}$$

Squaring both sides of this equation, $$(x-h)^2 + (y-k)^2 = r^2$$

This is called the standard equation of a circle. $(h,k)$ is the centre, $O$, and $r$ is the radius. You are required to memorize this equation and know how to apply it.

Another variant of this equation you are required to know and apply is $$x^2 + y^2 + 2fx + 2gy + c = 0$$ where $f, g, c \in \mathbb{R}$. Let me explain how to relate both variants to each other below.

By squaring the left-hand side of the standard equation of a circle then moving $r^2$ to the left-hand side, we obtain $$ \begin{align*} x^2 - 2hx + h^2 + y^2 - 2ky + k^2 &= r^2 \\ x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 &= 0 \end{align*} $$

Since $x^2 + y^2 + 2fx + 2gy + c = 0$ is merely another form of the same circle, meaning that both forms are geometrically identical to each other, let this equation be equivalent to $x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = 0$. $$x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 \equiv x^2 + y^2 + 2fx + 2gy + c$$

Now, compare the coefficients of like terms on both sides of the equation.

Equating the coefficients of $-2hx$ and $2fx$, $$ \begin{align*} -2h &= 2f \\ h &= -f \\ \end{align*} $$

Equating the coefficients of $-2ky$ and $2gy$, $$ \begin{align*} -2k &= 2g \\ k &= -g \\ \end{align*} $$

Equating the constant terms on both sides ($h^2 + k^2 - r^2$ and $c$), $$ \begin{align*} c &= h^2 + k^2 - r^2 \\ r^2 &= h^2 + k^2 - c \\ r &= \sqrt{h^2 + k^2 - c} \end{align*} $$

Since the radius is a distance which is a scalar quantity, we only take the principal square root (and ignore the $\pm$ sign before the $\sqrt{}$ symbol).

Recall that $h = -f$ and $k = -g$. Then, $$ \begin{align*} r &= \sqrt{(-f)^2 + (-g)^2 - c} \\ r &= \sqrt{f^2 + g^2 - c} \end{align*} $$

We have now derived three crucial formulae which will help you find the centre of the circle and the radius if given an equation of the form $x^2 + y^2 + 2fx + 2gy + c = 0$. $$ \begin{align*} h &= -f \\ k &= -g \\ r &= \sqrt{f^2 + g^2 - c} \end{align*} $$

Memorize these formulae as they will prove to be quite useful in questions.

Alternatively, if you prefer a more practical approach less reliant on memorization, you can convert to the standard form by completing the square. Once again, consider the equation $$x^2 + y^2 + 2fx + 2gy + c = 0$$ By grouping the $x$ terms separate from the $y$ terms, $$x^2 + 2fx + y^2 + 2gy + c = 0$$ Now, complete the square. $$ \begin{align*} (x+f)^2 - f^2 + (y+g)^2 - g^2 + c &= 0 \\ (x+f)^2 + (y+g)^2 - f^2 - g^2 + c &= 0 \\ (x+f)^2 + (y+g)^2 &= f^2 + g^2 - c \end{align*} $$

Although, at a first glance, it may look unfamiliar, we have converted the original equation to the form $(x-h)^2 + (y-k)^2 = r^2$, where $$ \begin{align*} h &= -f \\ k &= -g \\ r^2 &= f^2 + g^2 - c \end{align*} $$

Example 1: A circle is defined by the equation $x^2 + y^2 + 15x - 2y = 17$. Determine the centre of the circle and the length of its radius. Hence, express the circle in the form $(x-h)^2 + (y-k)^2 = r^2$.

Firstly, let us move the $17$ to the left-hand side to convert the equation to the form of $x^2 + y^2 + 2fx + 2gy + c = 0$. $$x^2 + y^2 + 15x - 2y - 17 = 0$$ By comparing the coefficients of the given equation to the specific form, we get $$2f = 15 \implies f = \frac{15}{2}$$ $$2g = -2 \implies g = -1$$ $$c = -17$$

Now, using the formula we previously derived, $$h = -f \implies h = -\frac{15}{2}$$ $$k = -g \implies k = 1$$ $$ \begin{align*} r = \sqrt{f^2 + g^2 - c} \implies r &= \sqrt{\left(-\frac{15}{2}\right)^2 + 1^2 - (-17)} \\ &= \sqrt{\frac{225}{4} + 1 + 17} \\ &= \sqrt{\frac{297}{4}} \\ r &= \frac{\sqrt{297}}{2} \end{align*} $$ We have found the values of $h$, $k$ and $r$. Substituting it into the equation $(x-h)^2 + (y-k)^2 = r^2$ now, $$ \begin{align*} \left[x - \left(-\frac{15}{2}\right)\right]^2 + (y-1)^2 &= \left(\frac{\sqrt{297}}{2}\right)^2 \\ \left(x + \frac{15}{2}\right)^2 + (y-1)^2 &= \frac{297}{4} \end{align*} $$

Example 2: A circle is defined by the equation $(x-3)^2 + (y+5)^2 = 100$. Determine the values of the constants $f$, $g$ and $c$ for which $x^2 + y^2 + 2fx + 2gy + c = 0$.

Conveniently, the given equation is already in the desired form of $(x-h)^2 + (y-k)^2 = r^2$. By simple observation, we can deduce that $h = 3$, $k = -5$ and $r^2 = 100 \implies r = 10$. All that needs to be done now is apply the formulae.

To find the value of $f$, $$ \begin{align*} h = -f \implies f &= -h \\ f &= -3 \end{align*} $$

To find the value of $g$, $$ \begin{align*} k = -g \implies g &= -k \\ g &= 5 \end{align*} $$

To find the value of $c$, first recall that $$r = \sqrt{f^2 + g^2 - c}$$

Squaring both sides, $$ \begin{align*} r^2 &= f^2 + g^2 - c \\ c &= f^2 + g^2 - r^2 \\ \implies c &= (-3)^2 + 5^2 - 10^2 \\ c &= 9 + 25 - 100 \\ c &= -66 \end{align*} $$

Finally, we can directly substitute in these values into $x^2 + y^2 + 2fx + 2gy + c = 0$. $$ \begin{align*} x^2 + y^2 + 2(-3)x + 2(5)y - 66 &= 0 \\ x^2 + y^2 - 6x + 10y - 66 &= 0 \end{align*} $$

Example 3: A circle, $C$, of radius $4$ passes through the point $(-1,-4)$ and has a centre $C$ is $(3,k)$. State the Cartesian equation of $C$ in the form $(x-h)^2 + (y-k)^2 = r^2$.

Right off the bat, we can substitute $h=3$ and $r=4$ into the equation to get $$ \begin{align*} (x-3)^2 + (y-k)^2 &= 4^2 \\ (x-3)^2 + (y-k)^2 &= 16 \end{align*} $$

Next, let us substitute $x=-1$ and $y=-4$ into the equation. $$ \begin{align*} (-1-3)^2 + (-4-k)^2 &= 16 \\ (-4)^2 + k^2 + 8k + 16 &= 16 \\ 16 + k^2 + 8k + 16 &= 16 \\ k^2 + 8k + 16 &= 0 \\ (k + 4)^2 &= 0 \\ k + 4 &= 0 \\ \implies k &= -4 \end{align*} $$

Finally, substituting $k=-4$ into the equation of the circle, $$ \begin{align*} (x-3)^2 + \left[y-(-4)\right]^2 &= 16 \\ (x-3)^2 + (y+4)^2 &= 16 \end{align*} $$

Tangent and Normal Lines to a Circle

A line is tangent to a circle if and only if it touches the circle at exactly one point.

A line $l$ is normal to a circle at a point if and only if the angle formed between the tangent line at the same point and $l$ is $\frac{\pi}{2}$ radians or $90^\circ$. $l$, normal to the circle at a point, passes through both that point on the circumference and the centre of the circle.

Consider a circle $C$ with equation $$x^2 + y^2 + 2fx + 2gy + c = 0 \tag{1}$$ Consider a line $l$ with equation $$y = ax + b \tag{2}$$ where $a$ is the gradient and $b$ is the $y$-intercept.

To determine the points of intersection between the line and the circle, you must solve the system of simultaneous equations above by directly substituting $y = ax + b$ into $(1)$.

Upon doing so, you will get a quadratic equation in terms of $x$. To determine the value of $a$ or $b$ for which

• $l$ is tangent to $C$, set the discriminant of the quadratic equation equal to $0$ and solve.
• $l$ intersects $C$ at two points, set the discriminant of the quadratic equation to be greater than $0$ and solve.
• $l$ does not intersect $C$ at all, set the discriminant of the quadratic equation to be less than $0$ and solve.

Recall that the product of the gradients of perpendicular lines is always $-1$. By definition, the tangent and normal lines are perpendicular to each other. Therefore, if $m_\text{tan}$ is the gradient of the tangent line and $m_\perp$ is the gradient of the normal line, then $$ \begin{align*} m_\text{tan} \times m_\perp &= -1 \\ m_\perp &= -\frac{1}{m_\text{tan}} \end{align*} $$

Example 4: The points $M(3,2)$ and $N(-1,4)$ are the ends of a diameter of circle $C$.

i) Determine the equation of circle $C$.

ii) Find the equation of the tangent to the circle at the point $P(-1,6)$.

Source: Caribbean Examinations Council, CSEC, 2016, Paper 2, Q3

i) The centre of $C$ is the midpoint of the diameter. Let $O(h,k)$ be the centre of $C$. Using the formula for the midpoint of a line segment, $$ \begin{align*} h &= \frac{3 + (-1)}{2} \\ &= \frac{3 - 1}{2} \\ &= \frac{2}{2} \\ h &= 1 \\ k &= \frac{2 + 4}{2} \\ &= \frac{6}{2} \\ k &= 3 \\ \end{align*} $$ Therefore, $O(1,3)$ is the centre of $C$.

ii) Let the equation of the tangent at $P$ be $y = ax + b$.

Recall the relationship between the gradient of the tangent and the gradient of the normal. $$ \begin{align*} m_\perp &= -\frac{1}{m_\text{tan}} \\ \implies m_\text{tan} &= -\frac{1}{m_\perp} \end{align*}$$

Remember that the centre of the circle and any point on the circumference lie on the normal line passing through that point. Therefore, by using the gradient of a line formula, the gradient of the normal line can be calculated. Recall $$m = \frac{y_2-y_1}{x_2-x_1}$$ Using the points $P(-1,6)$ and $O(1,3)$, $$ \begin{align*} m_\perp &= \frac{3-6}{1-(-1)} \\ &= \frac{-3}{1+1} \\ &= \frac{-3}{2} \\ m_\perp &= -\frac{3}{2} \end{align*} $$ Thus, $$ \begin{align*} m_\text{tan} &= -\frac{1}{-\frac{3}{2}} \\ m_\text{tan} &= \frac{2}{3} \\ \end{align*} $$

The equation of our tangent line is of the form $$y = \frac{2}{3}x + b$$

To find the value of $b$, substitute $x = -1$ and $y = 6$. $$ \begin{align*} 6 &= \frac{2}{3}(-1) + b \\ b &= 6 + \frac{2}{3} \\ b &= \frac{20}{3} \end{align*} $$ Finally, the equation of our tangent line is $$y = \frac{2}{3}x + \frac{20}{3}$$

Example 5: A circle $C$ is defined by the equation $(x−3)^2 + (y+1)^2 = 25$. A line $l$ is defined by the equation $y=\lambda x+4$ where $\lambda$ is a constant. Determine the value(s) of $\lambda$ such that $l$ is tangential to $C$ and hence deduce the equation(s) of $l$.

Firstly, define a system of equations representing the circle and the tangent line: $$(x−3)^2 + (y+1)^2 = 25 \tag{1}$$ $$y = \lambda x + 4 \tag{2}$$

Substituting $y = \lambda x + 4$ directly into $(1)$, $$ \begin{align*} (x - 3)^2 + (\lambda x + 4 + 1)^2 &= 25 \\ x^2 - 6x + 9 + (\lambda x + 5)^2 - 25 &= 0 \\ x^2 - 6x + 9 + \lambda^2x^2 + 10\lambda x + 25 - 25 &= 0 \\ x^2 + \lambda^2x^2 + 10\lambda x - 6x + 9 &= 0 \\ (\lambda^2 + 1)x^2 + (10\lambda - 6)x + 9 &= 0 \end{align*} $$

Setting the discriminant of this equation equal to 0 and solving for $\lambda$, $$ \begin{align*} (10\lambda - 6)^2 - 4(\lambda^2 + 1)(9) &= 0 \\ 100\lambda^2 - 120\lambda + 36 - 36(\lambda^2 + 1) &= 0 \\ 100\lambda^2 - 120\lambda - 36\lambda^2 - 36 &= 0 \\ 64\lambda^2 - 120\lambda &= 0 \\ 8\lambda^2 - 15\lambda &= 0 \\ \lambda(8\lambda - 15) &= 0 \\ \implies \lambda &= 0, \frac{15}{8} \end{align*} $$

In conclusion, the equation of $l$ can be either $y = 4$ or $y = \frac{15}{8}x + 4$. Both values of $\lambda$ make $l$ tangent to $C$ at a point.