Tangent to a Circle

In geometry, the concept of a tangent to a circle is essential for understanding geometric relationships and solving various problems involving circles. The tangent to a circle at a given point is a straight line that touches the circle at exactly one point without crossing it. This tangent line is perpendicular to the radius of the circle at the point of tangency. The equation of the tangent to a circle can be derived and expressed in several forms depending on the given information.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas  | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics 

Equation of the tangent to a Circle

A circle is the locus of a moving point such that its distance from a fixed point is constant.

The fixed point is called the center (O) of the circle and the constant distance is called its radius (r)

If the line L touches the circle, then Equation (iii) will have two equal real roots

So, Discriminant of equation (iii) = 0

B2−4AC=04 m2c2−4(1+m2)(c2−a2)=0a2=c21+m2c2=a2(1+m2)

In this case, the line is tangent to the circle

This is also the condition of tangency to the circle.

Equation of the Tangent in Point Form

The equation of the tangent to a circle x2+y2+2gx+2fy+c=0 at the point P(x1,y1) is x1+y1+g(x+x1)+f(y+y1)+c=0

Proof:

C(−g,−f) is the centre of the circle
As point P(x1,y1) lies on the circle.
∴ Slope of CP=y1−(−f)x1−(−g)=y1+fx1+g
Here, PT is the perpendicular to CP .
Thus, slope of PT =−(x1+gy1+f)
Hence, the equation of the tangent at P(x1,y1) is
now add gx1+fy1+c both side, we get

⇒xx1+yy1+g(x+x1)+f(y+y1)+c=x12+y12+2gx1+2fy1+c

i.e. x1+yy1+g(x+x1)+f(y+y1)+c=0
(As, point P(x1,y1) lies on the circle so, x12+y12+2gx1+2fy1+c=0 )

(y−y1)=−(x1+gy1+f)(x−x1)⇒(y−y1)(y1+f)+(x1+g)(x−x1)=0⇒xx1+yy1+gx+fy=x12+y12+gx1+fy1

Note:

In order to find out the equation of a tangent to any 2nd-degree curve, the following points must be kept in mind:

x2 is replaced by xx1
y2 is replaced by yy1
xy is replaced by xy1+x1y2
x is replaced by x+x122
y is replaced by y+y12

and c will remain c.

This method is applicable only for a 2nd degree conic.

Equation of Tangent of Circle in Parametric Form

The equation of the tangent at the point (acos⁡θ,asin⁡θ) to a circle x2+y2=a2 is xcos⁡θ+ysin⁡θ=a
Proof:
If S=x2+y2−a2=0 is the circle, then the tangent at (x1,y1) is T1= xx1+yy1−a2=0
put, x1=acos⁡θ,y1=asin⁡θ
we get, xcos⁡θ+ysin⁡θ=a

Equation of the Tangent in Slope Form

The equation of the tangent to a circle x2+y2=a2 having slope m is y=mx±a(1+m2), and point of tangency is (±am(1+m2),∓a(1+m2)).
Let y=mx+c be a tangent to the circle x2+y2=a2.
∴ Length of perpendicular from centre of circle (0,0)
on (y=mx+c)= radius of circle

∴|c|1+m2=a⇒c=±a1+m2

substituting this value of c in y=mx+c, we get y=mx±a(1+m2)
which are the required equations of tangents.

Corollary: It also follows that y=mx+c is tangent to x2+y2=a2 if c2=a2(1+m2) which is the condition of tangency.

Point of Contact:

Solving x2+y2=a2 and y=mx±a1+m2, simultaneously we get,

x=±am(1+m2)y=∓a(1+m2)

Thus, the coordinates of the points of contact are

(±am(1+m2),∓a(1+m2))

NOTE:
Equation of tangent of the circle (x−h)2+(y−k)2=a2 in terms of slope is (y−k)=m(x−h)±a(1+m2)

Recommended Video Based on Equation of the Tangent to a Circle

Solved Examples Based on Equation of the tangent to a Circle:

Example 1: Let the tangent to the circle x2+y2=25 at the point R(3,4) meet x-axis and y-axis at points P and Q , respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ , then r2 is equal to :
1) 52964
2) 58566
3) 12572
4) 62572

Solution

Tangent to the circle x² + y² = 25 at R(3, 4) is 3x + 4y = 25

I≡(62512254+253+12512,62512254+253+12512)∴I≡(62575+100+125,62575+100+125)≡(2512,2512)∴r2=(2512)2+(2512)2=62572

Example 2: Find the equation to the tangent of the circle x2+y2=26 at the point (5,1) :
1) x+5y=26
2) 5x+y=26
3) x+5y=26
4) 5x+y=26

Solution
Here x1=5,y1=1
So the equation of a tangent to a given circle at (5,1) is

xx1+yy1=26

Thus, we get 5x+y=26
Hence, the answer is the option (2).

Example 3: The tangent to the circle C1:x2+y2−2x−1=0 at the point (2,1) cuts off a chord of length 4 from a circle C2 whose centre is ( 3,−2 ). The radius of C2 is.
1) 2
2) 2
3) 3
4) 6

Solution
The equation of tangent on C1 at (2,1) is

xx1+yy1−(x+x1)−1=02x+y−(x+2)−1=0x+y=3

It cuts off the circle C2;
Distance of the line from the centre (3,−2) of C2

⇒|3−2−32|=2

Length of chord =4

Using concepts of intercepts

r2=( chord length/2 )2+( distance of line from centre )2⇒r2=4+2=6r=6

Example 4: A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If d₁ and d₂ are the distances of the tangent to the circle at the origin O from the points A and B respectively, then the diameter of the circle is:

1) 2d1+d−22
2) d1+2d22
3) d1+d2
4) d1d2d1+d2

Solution
Condition of tangency -
Length of perpendicular from centre of circle (0,0) on the line y=mx+c
is Radius of circle

i.e., |c|1+m2=ac=±a1+m2

- wherein

If y=mx+c is a tangent to the circle x2+y2=a2
Equation of circum circle of triangle OAB

x2+y2−ax−by=0
Equation of tangent at origin ax + by = 0.

d1=|a2|a2+b2 and d2=|b2|a2+b2d1+d2=a2+b2

= diameter

Example 5: A circle of radius r such that both coordinates of its centre are positive, touches the x -axis and line 3y−4x=0, then the equation of circle is
1) x2+y2+4rx−2ry+4r2=0
2) x2+y2−4rx−ry+4r2=0
3) x2+y2−2rx−2ry+r2=0
4) x2+y2−4rx−2ry+4r2=0

Solution

Answer (4)

Circle touches x -axis so its y -co-ordinate of x− axis = radius =r Now it touches line 3y−4x=0
So,

3r−4h5=±r⇒3r−4h=5r or 3r−4h=−5r⇒h=−2r4h=2r

h should be positive
So, h=2r
Now equation circle is

(x−2r)2+(y−r)2=r2x2+y2−4rx−2ry+4r2=0

Summary

The equation of the tangent to a circle is a fundamental concept in analytic geometry that provides insights into the geometric properties and spatial relationships involving circles. Whether deriving the tangent from an external point or at a specific point on the circle, understanding these equations is essential for solving geometric problems and applying these concepts in various fields, including optimization, computer graphics, and engineering design. By mastering the methods for finding tangents, one can effectively analyze and utilize the properties of circles in both theoretical and practical applications.