Equation of Tangent to a Circle – Condition of Tangency

When a straight line meet a circle on two coincident points then it is called the tangent to a circle. Here, you will learn condition of line to be a tangent  to a circle and equation of tangent to a circle with example.

Condition of Tangency :

The line L = 0 touches the circle S = 0 if P the length of the perpendicular from the center to that line and radius of the circle r are equal i.e.

P = r

Equation of Tangent to a Circle Formula

(i)  The Tangent at a point (\(x_1,y_1\)) on the circle \(x^2\) + \(y^2\) = \(a^2\) is

\(xx_1 + yy_1\) = \(a^2\)

(ii) The Tangent at the point (acost, asint) on the circle \(x^2\) + \(y^2\) = \(a^2\) is

xcost + ysint = a

The point of intersection of the tangents at the points \(P(\alpha)\) and \(Q(\beta)\) is (\(acos{{\alpha + \beta}\over 2}\over cos{{\alpha – \beta}\over 2}\), \(asin{{\alpha + \beta}\over 2}\over cos{{\alpha – \beta}\over 2}\)).

(iii) The equation of tangent at the points (\(x_1,y_1\)) on the circle \(x^2 + y^2 + 2gx + 2fy + c\) = 0 is

\(xx_1\) + \(yy_1\) + \(g(x + x_1)\) + \(f(y + y_1)\) + c = 0

(iv)  If line y = mx + c is a straight line touching the circle \(x^2\) + \(y^2\) = \(a^2\), then

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

and contact points are

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

and the equation of tangent is

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

(iv) The equation of tangent with slope m of the circle \((x-h)^2 + (y-k)^2\) = \(a^2\) is

(y – k) = m(x – h) \(\pm a\sqrt{1 + m^2}\)

Example : Find the tangent to the circle \(x^2 + y^2 – 2ax\) = 0 at the point (5, 6).

Solution : Since the tangent at the points (\(x_1,y_1\)) on the circle \(x^2 + y^2 + 2gx + 2fy + c\) = 0 is \(xx_1\) + \(yy_1\) + \(g(x + x_1)\) + \(f(y + y_1)\) + c = 0.

5x + 6y – a(x + 5) = 0

\(\implies\) 5x + 6y – ax – 5a = 0
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