Table of Contents
Last modified on August 3rd, 2023
The Latin word ‘tangent’ means, ‘to touch’. The tangent of a circle is defined as a straight line that touches the circle at a single point. The point where the tangent touches the circle is called the ‘point of tangency’ or the ‘point of contact’. At the point of tangency, a tangent is perpendicular to the radius of the circle.
Shown below is a tangent to the given circle.
Let’s learn some of the common rules of the tangent line to a circle.
A line or a line segment is considered a tangent only when it touches a curve at a single point or else it is simply a line or a line segment. Thus, based on the point of tangency and its location with respect to the circle, there can be three possible conditions for the tangent as given below:
If ‘P’ is a point inside the circle, all the lines through point ‘P’ intersect the circle at two points. Thus, no tangent can be drawn to a circle, which passes through a point lying inside the circle.
If ‘P’ is a point on the circle, then from the given figure, we can say that there is only one tangent possible through a given point on the circle.
If ‘P’ is a point outside the circle, then from the given figure, we can say that there are exactly two tangents to circle from a point which lies outside the circle.
The formula to calculate the tangent of a circle is derived below:
Suppose a point ‘T’ lies outside the circle. From that point, we will draw two tangents to the circle. In the given figure, if ‘RT’ is the secant and ‘QT’ is the tangent of the circle, then the relation between the secant and the tangent is given as:
RT/QT = QT/PT
By cross multiplying, we get,
QT2 = RT × PT
Let’s work out a few example problems involving the tangent of a circle.
Given that OP = 4 units, PR= 3 units. Find the length of OR.
In the given figure, PR is the tangent and OP is the radius
Now, by tangent radius theorem, we know,
PR⊥OR, which makes triangle OPR a right triangle
Thus, applying the Pythagorean Theorem we get,
(OR)2 = (OP)2 + (PR)2
=> (OR)2 = (4)2 + (3)2
=> (OR)2 = 16 + 9 = 25 units
OR = 5 units
Find the length of the tangent in the circle shown alongside.
In the given figure, there is one tangent and one secant
Given that, PQ = 5 cm, QR = 15 cm
Therefore,
PR = PQ + QR = (5 + 15) = 20 cm
Now, according to the formula of the tangent of a circle,
SR2 = PR × QR
SR2 = 20× 15 = 300 = 17.32 cm
Find the length of the tangent PR if the radius of the given circle is 6 m. (Given that OR = 10 m).
Given that, PR is the tangent to the given circle; it is perpendicular to the radius OP
Thus, applying Pythagorean Theorem we get,
(OR)2 = (OP)2 + (PR)2, here OR = 10 m, OP = 6 m
(PR)2 = (OR)2 – (OP)2
=> (PR)2 = (10)2 – (6)2
=> (PR)2 = 100 – 36 = 64
=> PR = 8 m
Thus, the length of the tangent PR = 8 m
The standard form of the equation of a circle with center at (h, k), radius r, and a point L (x, y) on the circumference of the circle is given by:
r2 = (x-h)2 + (y-k)2
Let us learn the steps to find the equation of the tangent to a circle using the above formula.
In order to determine the equation of a tangent, follow the following steps:
1) Determine the equation of the circle and write it in the form:
r2 = (x-h)2 + (y-k)2
2) From the equation, determine the coordinates of the center of the circle (h, k)
3) Determine the slope of the radius of the circle by using the formula:
mOL = y2 – y1/x2 – x1
4) The radius OP being perpendicular to the tangent PQ, it is written as
mPQ =-1/mOL
5) Write the slope-intercept form of the equation of a circle substituting the value of mPQ and the coordinates of P, making ‘y’ the subject of the equation:
y – y1 = m(x – x1)
Let’s work out a problem to clear your concept better.
Determine the equation of the tangent to the circle x2 + y2 + 8x + 2y – 27 = 0 at point (-3, 2).
As we know,
The standard form of the equation of a circle with center at (h, k), radius r, and a point L (x, y) on the circumference of the circle is given by:
r2 = (x-h)2 + (y-k)2
The given equation of the circle is:
x2 + y2 + 8x + 2y – 27 = 0
To write the equation in the form of:
r2 = (x-h)2 + (y-k)2, with center at (h, k), radius r, and a point L (x, y) on the circumference of the circle
x2 + y2 + 8x + 2y – 27 = 0
x2 + y2 + 8x + 2y = 27
(x2 + 8x + 16) + (y2 + 2y + 1) = 27 – 17
(x + 4)2 + (y + 1)2 = 10
Thus, the center of the circle is at (-4, -1) and the radius is √10
Plotting the circle with center at (-4, -1) and radius √10
Now, we will determine the gradient of the radius
Determining the Gradient of the Radius mOL
Then,
mOL = y2 – y1/x2 – x1
Let (x1, y1) = (-4, -1), and (x2, y2) = (-3, 2)
Thus,
mOL = 2 – (-1)/-3 – (-4)
= 3
Determining the Gradient of the Tangent mPQ
As we know,
mOL × mPQ = -1
=> 3 × mPQ = -1
=> mPQ = -1/3
Determining the Equation of the Tangent to the Circle
Sunstituting the value of the slope of the tangent in the slope-intercept form of the the equation of a circle:
y – y1 = m(x – x1)
Here, m = mPQ = -1/3 and (x1, y1) = (-3, 2)
Substituting the values in the equation we get,
y – 2 = -1/3(x – (-3))
=> y – 2 = -1/3(x + 3)
=> y – 2 = 1/3x – 1
=> y = 1/3x – 1 + 2
=> y = 1/3x + 1
Thus, the equation of the tangent to the Circle is y = 1/3x + 1
Last modified on August 3rd, 2023