How to Find Tangent and Normal to a Circle

In geometry, tangents and normals have great importance. A straight line that touches a circle at only one point is the tangent of the circle. The normal to a circle is a straight line drawn at 90 to the tangent at the point where the tangent touches the circle. In this article, we will discuss how to find the tangent and normal to a circle.

Steps to find Tangent and Normal to a Circle

Step 1. If x2 + y2 = a2 is a circle, then

a. The equation of a tangent to the circle at (x1, y1) is given by xx1 + yy1 = a2.

b. The equation of normal to the circle at (x1, y1) is given by yx1 – xy1 = 0.

c. The equation of a tangent to the circle at (a cos θ, a sin θ) is given by x cos θ + y sin θ = a.

d. The equation of a normal to the circle at (a cos θ, a sin θ) is given by x sin θ – y cos θ = 0.

Step 2. If the circle is given by x2 + y2 + 2gx + 2fy + c = 0

a. The equation of a tangent to the circle at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0

b. The equation of normal to the circle at (x1, y1) is (y – y­­1)/(y1 + f) = (x – x1)/(x1 + g).

Step 3. For a line y = mx + c to be a tangent to a circle x2 + y2 = a2, it should satisfy c = ± a(√(1+m2). The equation of tangent is given by y = mx ± a(√(1+m2).

Particular Cases of Circle

Solve Examples:

 Example 1: The tangent to circle [latex]{{x}^{2}}+{{y}^{2}}=5[/latex] at (1, -2) also touches the circle [latex]{{x}^{2}}+{{y}^{2}}-8x+6y+20=0.[/latex] Find the coordinates of the corresponding point of contact.

Solution:

Equation of tangent to [latex]{{x}^{2}}+{{y}^{2}}=5[/latex] at (1, -2) is

[latex]x\left( 1 \right)+y\left( -2 \right)-5=0[/latex]

or [latex]x-2y-5=0[/latex]

putting [latex]x=2y+5[/latex] in second circle, we get

[latex]{{\left( 2y+5 \right)}^{2}}+{{y}^{2}}-8\left( 2y+5 \right)+6y+20=0[/latex] [latex]\Rightarrow \,\,{{y}^{2}}+2y+1=0[/latex] [latex]\Rightarrow \,\,y=-1[/latex] [latex]\Rightarrow \,\,x=-2+5=3[/latex]

Thus, point of contact is (3, -1).

Example 2: Find the angle between the two tangents from the origin to the circle [latex]{{\left( x-7 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=25.[/latex]

Solution:

Any line through (0, 0) be [latex]y-mx=0[/latex] and it is a tangent to circle [latex]{{\left( x-7 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=25.[/latex]

If [latex]\frac{\left| -1-7m \right|}{\sqrt{1+{{m}^{2}}}}=5[/latex] [latex]\Rightarrow m=\frac{3}{4},-\frac{4}{3}[/latex]

The product of both the slopes is -1.

Hence, the angle between the two tangents is [latex]\pi /2.[/latex]

Example 3: Find the equation of normal to the circle [latex]2{{x}^{2}}+2{{y}^{2}}-2x-5y+3=0[/latex] at (1, 1).

Solution:

The centre of the circle is (1/2, 5/4)

Normal to circle at point (1, 1) is line passing through the points (1, 1) and (1/2, 5/4) which is [latex]x+2y=3.[/latex]

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