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 $$\begin{array}{rl}{x}^2-2hx+h^2+y^2-2ky+k^2& =r^2\\ x^2+y^2-2hx-2ky+h^2+k^2-r^2& =0\end{array}$$

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 $2fx$, $$\begin{array}{rl}-2h& =2f\\ h& =-f\end{array}$$

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

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

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, $$\begin{array}{rl}r& =\sqrt{(-f{)}^2+(-g{)}^2-c}\\ r& =\sqrt{f^2+g^2-c}\end{array}$$

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 $x^2+y^2+2fx+2gy+c=0$. $$\begin{array}{rl}h& =-f\\ k& =-g\\ r& =\sqrt{f^2+g^2-c}\end{array}$$

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. $$\begin{array}{rl}(x+f{)}^2-f^2+(y+g{)}^2-g^2+c& =0\\ (x+f{)}^2+(y+g{)}^2-f^2-g^2+c& =0\\ (x+f{)}^2+(y+g{)}^2& =f^2+g^2-c\end{array}$$

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 $$\begin{array}{rl}h& =-f\\ k& =-g\\ r^2& =f^2+g^2-c\end{array}$$

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 $$2f=15⟹f=\frac{15}{2}$$ $$2g=-2⟹g=-1$$ $$c=-17$$

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

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 $r^2=100⟹r=10$. All that needs to be done now is apply the formulae.

To find the value of $f$, $$\begin{array}{rl}h=-f⟹f& =-h\\ f& =-3\end{array}$$

To find the value of $g$, $$\begin{array}{rl}k=-g⟹g& =-k\\ g& =5\end{array}$$

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

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

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

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 $$\begin{array}{rl}(x-3{)}^2+(y-k{)}^2& =4^2\\ (x-3{)}^2+(y-k{)}^2& =16\end{array}$$

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

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

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 $$\begin{array}{}\text{(1)}& x^2+y^2+2fx+2gy+c=0\end{array}$$ Consider a line $l$ with equation $$\begin{array}{}\text{(2)}& y=ax+b\end{array}$$ 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 $$\begin{array}{rl}{m}_{\text{tan}}\times m_{\perp }& =-1\\ m_{\perp }& =-\frac{1}{m_{\text{tan}}}\end{array}$$

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, $$\begin{array}{rl}h& =\frac{3+(-1)}{2}\\ & =\frac{3-1}{2}\\ & =\frac{2}{2}\\ h& =1\\ k& =\frac{2+4}{2}\\ & =\frac{6}{2}\\ k& =3\end{array}$$ 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. $$\begin{array}{rl}{m}_{\perp }& =-\frac{1}{m_{\text{tan}}}\\ ⟹m_{\text{tan}}& =-\frac{1}{m_{\perp }}\end{array}$$

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 $O(1,3)$, $$\begin{array}{rl}{m}_{\perp }& =\frac{3-6}{1-(-1)}\\ & =\frac{-3}{1+1}\\ & =\frac{-3}{2}\\ m_{\perp }& =-\frac{3}{2}\end{array}$$ Thus, $$\begin{array}{rl}{m}_{\text{tan}}& =-\frac{1}{-\frac{3}{2}}\\ m_{\text{tan}}& =\frac{2}{3}\end{array}$$

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 $y=6$. $$\begin{array}{rl}6& =\frac{2}{3}(-1)+b\\ b& =6+\frac{2}{3}\\ b& =\frac{20}{3}\end{array}$$ 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: $$\begin{array}{}\text{(1)}& (x-3{)}^2+(y+1{)}^2=25\end{array}$$ $$\begin{array}{}\text{(2)}& y=\lambda x+4\end{array}$$

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

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

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.