# Equation of Tangent & Normal on the Circle

### Definition of tangent:

The tangent at any point on a circle is defined to be a straight line which meets the circle at that point but being produced, does not cut it.

The point where tangent meets the curve is called point of contact.

In case of circles, the tangent is always perpendicular to the radius drawn from the centre to the point of contact.

If S = 0 be a curve then S1 = 0 indicates the equation which is obtained by substituting
x = x1 and y = y1 in the equation of the given curve, and T = 0 is the equation which is obtained by substituting x2 = xx1 , y2 = yy1 ,

2xy = xy1 + yx1 , 2x = x + x1

and 2y = y + y1 in the equation S = 0.

If S ≡ x2 + y2 + 2gx + 2fy + c = 0

then S1 ≡ x12 +y12 + 2gx1 + 2fy1 + c

and T ≡ xx1 + yy1 + g( x + x1) + f(y + y1) + c, then

Equation of the tangent to x2 + y2 + 2gx + 2fy + c = 0 at A(x1, y1) is

xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0, then

The condition that the straight line y = mx + c is a tangent to the circle x2 + y2 = a2 is

c2 = a2 (1 + m2) and

the point of contact is (− a2m/c, a2/c)

i.e. y = mx ± a√(1+m2) is always a tangent to

the circle x2 + y2 = a2 whatever be the value of m.

Note:

A line will touch a circle if and only if the length of the perpendicular from the centre of the circle to the line is equal to the radius of the circle.

The joint equation of a pair of tangents drawn from the point A(x1, y1) to the circle  x2 + y2 + 2gx + 2fy + c = 0 is T2 = SS1

### Definition of Normal:

The normal at any point of a curve is the straight line which is perpendicular to the tangent at the point of tangency (point of contact).

So, in case of circles, normal always passes through the centre of the circle.

The equation of the normal to the circle x2 + y2 + 2gx + 2fy + c = 0 at any point (x1 , y1) lying on the circle is

$\large \frac{x – x_1}{x_1 + g} = \frac{y – y_1}{y_1 + f}$

Example :. Find the equation of a circle which touches the coordinate axes and the line x + y =1 and has its centre situated remote from the origin with respect to the given line.

Solution: Let r be the radius of the required circle. Since the circle both the axes. Its centre will lie on the line y = x.

Let centre C ≡ (r , r) . Also x + y = 1 is a tangent to the required circle.

=> perpendicular distance from (r, r) to

x + y − 1 = 0 is radius r.

=> |(r + r − 1)√2| = r

=> r = 1/(2 − √2) , 1/(2 + √2)

Since centre is situated remote from the origin,

$\large r = \frac{1}{2-\sqrt{2}}$

$\large C = (\frac{1}{2-\sqrt{2}} ,\frac{1}{2-\sqrt{2}})$

Therefore, the required equation is

$\large (x-\frac{1}{2-\sqrt{2}})^2 + (y-\frac{1}{2-\sqrt{2}})^2 = (\frac{1}{2-\sqrt{2}})^2$