Equation of Tangent to the Circle: The meaning of the word tangent is ‘to touch’. The Latin word for the same is ‘tangere’. In general, we can say that the line that touches a circle precisely at one point on the circumference. In geometry, a tangent is a line drawn outside the circle and touches the curve at one point.

A real-life example of a tangent is when you ride a bicycle, every point on the circumference of the wheel makes a tangent with the road. In this article, let us learn more about tangents, their properties, and the equation of the tangent to a circle.

Definition of a Tangent

A tangent is a line that passes through a point on a curve. The following figure shows an arc \(S\) and a point \(P\) on the curve.

A tangent at \(P\) has been drawn to \(S\).

Point of Tangency

The point of tangency is where the line touches the curve at one point. In the above figure, point \(P\) represents the point of tangency.

A tangent to the circle is the line that touches the circle at one point only. A circle can have only one tangent at a point on the circle. The point where the tangent comes in contact with the circle is called the point of tangency.

The following figure shows a circle with a point \(P\).

Based on where the point of tangency lies with respect to the circle, we can define the conditions for tangent as:

• When the point lies inside the circle
• When the point lies on the circle
• When the point lies outside the circle

When the Point Lies Inside the Circle

Observe that \(P\) lies inside the circle as shown in the figure. All the lines through point \(P\) intersect the circle at two distinct points. So, no tangent can be drawn to a circle that passes through a point lying inside the circle.

When the Point Lies on the Circle

Only one tangent can be drawn to the circle at \(P\) that lies on the circle. There is only one point of tangency.

When the Point Lies Outside the Circle

When \(P\) lies outside the circle, observe that two tangents can be drawn to the given circle. Hence, there are two points of tangency. Also, the length of tangents from the points of tangency to \(P\) are always equal.

Properties of Tangents

A tangent has the following properties:

• Tangent always touches the circle at one point only.
• Tangent is perpendicular to the radius \((r)\) of the circle at the point of tangency.

• Tangent never intersects the circle at more than one point.
• The length of tangents drawn to a circle from an external point are equal.

Formula for Tangents

The formula for tangents is as given below.

Let a point \(P\) lie outside the circle. From \(P\), we can draw two tangents to the circle at points of tangency \(A\) and \(B\).

Now, let a secant be drawn from the point \(P\) to intersect the circle at \(Q\) and \(R\). \(PS\) is the tangent from \(P\) to \(S\).

Formula for the tangent-secant of the circle is \(\frac{{PR}}{{PS}} = \frac{{PS}}{{PQ}}\)

\( \Rightarrow P{S^2} = PQ \times PR\)

What is Equation of a Circle?

A circle is a closed curve which is drawn from the fixed point called the centre, in which each point on the curve is having the same distance from the centre to any point on the centre. The equation of a circle with \(\left( {h,\,k} \right)\) centre and \(r\) radius is given by \({(x – h)^2} + {(y – k)^2} = {r^2}\), which is a Centre-Radius form of a circle.

The general equation of a circle is \({x^2} + {y^2} + 2gx + 2fy + c = 0\), where \(g,\,f\) and \(r\) are constants, here centre \( = ( – g,\, – f)\) and radius \(r = \sqrt {{g^2} + {f^2} – c} \).

What is Slope of a Tangent?

There are several ways in which you can find the slope of a tangent to a function. These include actually drawing a plot of the function and the tangent line and physically measuring the slope and also using successive approximations via secants. However, for simple algebraic functions, the quickest approach is to use calculus. The calculus method takes the derivative of the function at the point, which is equal to slope of the tangent at that point.

Condition for Tangency

A line touches the circle if the length of the perpendicular drawn from the centre of the circle from the line is equal to the radius of the circle.

A circle \({x^2} + {y^2} = {a^2}\) will touch the line \(y = mx + c\) if \(c = \pm a\sqrt {1 + {m^2}} \).

• If \({a^2}\left( {1 + {m^2}} \right) – {c^2} > 0\), then the line will meet the circle at two real and distinct points
• If \({c^2} = {a^2}\left( {1 + {m^2}} \right)\), then the line will touch the circle.
• If \({a^2}\left( {1 + {m^2}} \right) – {c^2} < 0\), then the line will meet the circle at two imaginary points.

Equation of Tangent to the Circle

Point Form

The equation of a tangent to the circle \({x^2} + {y^2} = {a^2}\) at the point \(\left( {{x_1},\,{y_1}} \right)\) is \(x{x_1} + y{y_1} = {a^2}\)

The equation of a tangent to the circle \({x^2} + {y^2} + 2gx + 2fy + c = 0\) at the point \(\left( {{x_1},\,{y_1}} \right)\) is given by,

\(x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0\)

Slope Form

The condition of tangency for a line \(y = mx + c\) to the circle \({x^2} + {y^2} = {a^2}\) is \(c = \pm a\sqrt {1 + {m^2}} \). The equation of a tangent to the circle \({x^2} + {y^2} = {a^2}\) is:

\(y = mx \pm a\sqrt {1 + {m^2}} \)

The point of contact is given by,

\(\left( {\frac{{ \mp am}}{{\sqrt {1 + {m^2}} }},\,\frac{{ \pm a}}{{\sqrt {1 + {m^2}} }}} \right)\)

Parametric Form

The equation of a tangent to the circle \({x^2} + {y^2} = {a^2}\) at the point \((a\,\cos \,\theta ,\,a\,\sin \,\theta )\) is given by,

\(x\,\cos \,\theta + y\,\sin \,\theta = a\)

Solved Examples – Equation of Tangent to the Circle

Q.1. If the line \(x + y = 1\) is a tangent to a circle with centre \((2,\,3)\), then its equation will be
Ans: The radius of the circle \( = \) Perpendicular distance of \((2,\,3)\) from the line \(x + y = 1\) is \(\frac{{|2 + 3 – 1|}}{{\sqrt {{1^2} + {1^2}} }} = \frac{4}{{\sqrt 2 }} = 2\sqrt 2 \)
Using centre-radius form: \({(x – h)^2} + {(y – k)^2} = {(r)^2}\)
Therefore, \({(x – 2)^2} + {(y – 3)^2} = 8\)
\( \Rightarrow {x^2} + {y^2} – 4x – 6y + 4 + 9 = 8\)
\( \Rightarrow {x^2} + {y^2} – 4x – 6y + 5 = 0\)
Therefore, the required equation of a circle is \({x^2} + {y^2} – 4x – 6y + 5 = 0\)

Q.2. The line \(3x – 2y = k\) meets the circle \({x^2} + {y^2} = 4{r^2}\) at only point, if \({k^2}=\)
Ans: Given: Equation of the line is \(3x – 2y = k\) 
circle is \({x^2} + {y^2} = 4{r^2}\)
Equation of a line can be written as \(y = \frac{3}{2}x – \frac{k}{2}\)
On comparing with \(y = mx + c\), we get:
\(c = – \frac{k}{2},\,m = \frac{3}{2}\)
The condition for the line \(y = mx + c\) to be a tangent to the circle \({x^2} + {y^2} = {a^2}\) is
\({c^2} = {a^2}\left( {1 + {m^2}} \right)\) or \(c = \pm a\sqrt {1 + {m^2}} \)
Therefore, \(\frac{{ – k}}{2} = ( \pm 2r)\sqrt {1 + {{\left( {\frac{3}{2}} \right)}^2}} \)
\( \Rightarrow \frac{{{k^2}}}{4} = 4{r^2} \times \frac{{13}}{4}\)
\(\therefore \,{k^2} = 52{r^2}\)

Q.3. Find the equation of tangent to the circle \({x^2} + {y^2} – 2y + 6x – 7 = 0\) at the point \(P{\rm{( – 2,}}\,{\rm{5)}}\).
Ans: Given: Equation the circle is \({x^2} + {y^2} – 2y + 6x – 7 = 0\)
Equation of the tangent to the circle \({x^2} + {y^2} + 2gx + 2fy + c = 0\) at the point \(\left( {{x_1},\,{y_1}} \right)\) is
\(x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0\)
\( \Rightarrow x( – 2) + y(5) + 3(x – 2) – 1(y + 5) – 7 = 0\)
\( \Rightarrow – 2x + 5y + 3x – 6 – y – 5 – 7 = 0\)
\( \Rightarrow x + 4y – 18 = 0\)
Therefore, the required equation of the tangent is \(x + 4y – 18 = 0\).

Q.4. The equation of a tangent to the circle at the point \({\rm{(1,}}\,{\rm{ – 1)}}\), whose centre is the point of intersection of the following lines \(x – y = 1\) and \(2x + y = 3\) is 
Ans: Given lines:
\(x – y = 1 \to (1)\)
\(2x + y = 3 \to (2)\)
By adding equations \((1)\) and \((2)\), then we get:
\(3x{\rm{ = 4}}\)
\( \Rightarrow x = \frac{4}{3}\)
Substitute the value of \(x = \frac{4}{3}\)  in any one of the equations,
Therefore, \(y = \frac{1}{3}\)
Thus, the centre is \(\left( {\frac{4}{3},\,\frac{1}{3}} \right)\)

Now, the slope of \(OP\left( {{m_1}} \right) = \frac{{\frac{1}{3} + 1}}{{\frac{4}{3} – 1}}\)
\( = \frac{{\frac{4}{3}}}{{\frac{1}{3}}}\)
\( = 4 \)
Hence, the slope of the tangent is \({m_2} = – \frac{1}{4}\)  [Since the tangent is perpendicular to the radius of the circle at the point of tangency, in other words, two lines are perpendicular, then the product of their slopes is equal to \({m_1} \times {m_2} = – 1\)]
Equation of the tangent is \(y + 1 = – \frac{1}{4}(x – 1)\)
\( \Rightarrow 4y + 4 = – x + 1\)
\( \Rightarrow x + 4y + 3 = 0\)
Hence, the equation of the tangent is \(x + 4y + 3 = 0\)

Q.5. What is the equation of any tangent to the circle \({x^2} + {y^2} – 2x + 4y – 4 = 0\)?
Ans: Given circle: \({x^2} + {y^2} – 2x + 4y – 4 = 0\)
Here, centre \(C(1,\, – 2)\) and radius \( = \sqrt {1 + 4 + 4} = 3\)
Let \(y = mx + c\) be the tangent to the circle \({x^2} + {y^2} – 2x + 4y – 4 = 0\)
Thus, the distance of a line from the centre of the circle \({x^2} + {y^2} – 2x + 4y – 4 = 0\) and radius will be equal.
Therefore, \(\left| {\frac{{m + c + 2}}{{\sqrt {1 + {m^2}} }}} \right| = r = 3\)
\( \Rightarrow |m + c + 2| = 3\sqrt {1 + {m^2}} \)
\( \Rightarrow m + c + 2 = \pm 3\sqrt {1 + {m^2}} \)
\( \Rightarrow c = \pm 3\sqrt {1 + {m^2}} – (m + 2)\)
Hence, the required tangent is \(y = mx \pm 3\sqrt {1 + {m^2}} – (m + 2)\).

Q.6. Find the equation of the tangent to the circle \({x^2} + {y^2} = 25\) at the point \(P\left( { – 3,\,4} \right)\).
Ans: As we know, the equation of the tangent to the circle \({x^2} + {y^2} = {a^2}\) at the point \(P\left( {{x_1},\,{y_1}} \right)\) is \(x{x_1} + y{y_1} = {a^2}\)
So, \(x{\rm{( – 3) + y(4) = 25}}\)
\( \Rightarrow – 3x + 4y = 25\)
\( \Rightarrow 3x – 4y = – 25\)
\( \Rightarrow 3x – 4y + 25 = 0\)
Therefore, the required equation of the tangent is \(3x – 4y + 25 = 0\).

Summary

A tangent to the circle is the line that touches the circle at one point. A circle can have only one tangent at a point to the circle. This point where the tangent comes in contact with the circle is called the point of tangency. i.e., a line \(L{\rm{ = 0}}\) touches the circle \(S{\rm{ = 0}}\), if the length of the perpendicular drawn from the centre of the circle to the line is equal to the radius of the circle. Also, we learnt about how to find the equation of the tangent to the circle at a given point using different forms, i.e., point form, slope form and parametric form.

Frequently Asked Questions (FAQs)

Q1. How do you find the equation of a tangent to a circle?
Ans: The equation of a tangent to a circle can be determined by using point form, slope form and parametric form.

• Point form: \(x{x_1} + y{y_1} = {a^2}\)
• Slope form: \(y = mx \pm a\sqrt {1 + {m^2}}\)
• Parametric form: \(x\,\cos \,\theta + y\,\sin \,\theta = a\)

Q2. What is the slope of a tangent?
Ans: There are several ways to find the slope of a tangent to a function. These include actually drawing a plot of the function and the tangent line and physically measuring the slope, and for simple algebraic functions, the quickest approach is to use calculus.

Q3. How do you write the equation of a circle with the centre and tangent?
Ans: To find the equation of the circle, we need the centre and radius. Here, the radius is the perpendicular distance from the centre to the tangent. Then we can use these values centre and radius to find the equation of the circle.

Q4. How is a tangent line related to the radius of the circle?
Ans: When a radius of a circle is drawn from the centre to the point of tangency, it is perpendicular to the tangent line at that point.

Q5. What is the definition of tangent lines to circles?
Ans: A tangent line to a circle is a line that touches the circle at exactly one point, never entering the circle’s interior.
The below figure shows the relation between the tangent line and the circle, where the tangent line touches the circle exactly once.

Q6. What is the difference between tangent and normal?
Ans: A tangent to the curve is a line that touches the curve at one point. A normal to a curve is a line perpendicular to a tangent to the curve. The relationship between tangent and normal can be seen in the below figure:

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