# Tangents To Circles

We first consider the case which we’ve already considered in Example - 8 earlier, namely, when will the line \(y = mx + c\) be a tangent to the circle \({x^2} + {y^2} = {a^2}?\)

Earlier, we solved the simultaneous system of two equations and put the discriminant of the resulting quadratic equal to 0 to obtain the condition for tangency. Here, we follow an alternative approach.

We use the simple geometric fact that if a line *L* is a tangent to a circle *S*, then the perpendicular distance of the centre of *S* from *L* must equal the radius of *S*:

In the current case, the centre of \({x^2} + {y^2} = {a^2}\) is (0, 0) and its radius is *a*. The perpendicular distance of (0, 0) from \(y = mx + c\) must equal *a*, i.e.

\[\begin{align}&\frac{{\left| c \right|}}{{\sqrt {1 + {m^2}} }} = a \\\\ \Rightarrow \quad &\fbox{${{c^2} = {a^2}(1 + {m^2})}$}\end{align}\]

This is the same condition that we derived earlier. From this relation, we can infer that any line of the form \(y = mx \pm a\sqrt {1 + {m^2}} \) will always be a tangent to the circle \({x^2} + {y^2} = {a^2},\) whatever the value of *m* may be.

Consider now the problem of finding the equation of the tangent to the circle \({x^2} + {y^2} = {a^2}\) at a point \(P({x_1},{y_1})\) lying *on* the circle.

The tangent *T* will obviously be perpendicular to *OP*. Thus, the equation of *T* is

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{{y - {y_1}}}{{x - {x_1}}} = \frac{{ - {x_1}}}{{{y_1}}}\\\\ \Rightarrow \quad & x{x_1} + y{y_1} = x_1^2 + y_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\end{align}\]

The other piece of information that we have is that \(({x_1},{y_1})\) lies on the circle and therefore must satisfy its equation. Thus, \(x_1^2 + y_1^2 = {a^2}.\) Using this in (1), we obtain the equation of *T* as

\[\fbox{$\begin{array}{*{20}{c}} {T:\,\,x{x_1} + y{y_1} = {a^2}}\end{array}$}\]

If the point \(({x_1},{y_1})\) has been specified in polar form, i.e. in the form \((a\cos \theta ,a\sin \theta ),\) the equation of *T* becomes

\[\fbox{$\begin{array}{*{20}{c}} {T:\,\,x\cos \theta + y\sin \theta = a}\end{array}$}\]

We now go to the general case of finding the equation of the tangent to an arbitrary circle \(S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0\) at a point \(P({x_1}{y_1})\) lying on this circle:

Here again, as in the earlier case, we have two pieces of information which we can put to use :

* \(\begin{align} \text{ P lies on S} : \qquad \quad &x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c = 0 \qquad\qquad ...(1)\\\\\text{ T is perpendicular to CP} : &\frac{{y - {y_1}}}{{x - {x_1}}} \times \frac{{{y_1} + f}}{{{x_1} + g}} = - 1\,\qquad\left\{ {(x,y){\text{ is any point on }}T} \right\}\\\\ \Rightarrow \qquad &x{x_1} + y{y_1} + gx + fy = x_1^2 + y_1^2 + g{x_1} + f{y_1}\end{align}\)*

We now add \((g{x_1} + f{y_1} + c)\) on both sides above :

\( \Rightarrow \quad x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c\)

Now we use (1) for the *RHS *above to finally obtain the equation of *T* as

\(\qquad\qquad\qquad \fbox{$\begin{array}{*{20}{c}} {T:\,\,x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0}\end{array}$}\)

The expression on the left hand side of the equation above is conventionally denoted as \(T({x_1},{y_1})\) (Note that *T* is a function of *x* and *y* too). Thus, the equation of the tangent can be written concisely as

\[T({x_1},{y_1}) = 0 \quad : \quad \text {Equation of tangent at}\;\;({x_1},{y_1}) \; \; \text{to} \;\;S\]

Using an analogous approach, we can write the equation of the normal to the circle \({x^2} + {y^2} + 2gx + 2fy + c = 0\) at the point \(P\left( {{x_1},{y_1}} \right)\). You are urged to do this yourself. Note that every normal of a circle will pass through the circle’s centre.

**Example - 16**

Find the equation of the tangent to

(a) \({x^2} + {y^2} = 1\) at \(\begin{align}\left( {\frac{1}{{\sqrt 3 }},\frac{{\sqrt 2 }}{{\sqrt 3 }}} \right)\end{align}\)

(b) \({x^2} + {y^2} - 2x - 4y + 4 = 0\) at \(\begin{align}\left( {\frac{1}{2},2 + \frac{{\sqrt 3 }}{2}} \right)\end{align}\)

**Solution (a)** The equation of the tangent to \({x^2} + {y^2} = {a^2}\) at \(({x_1},{y_1})\) is, as obtained in the preceding discussion,

\[\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&T({x_1},{y_1}) = 0\\\\ \Rightarrow & x{x_1} + y{y_1} = {a^2}\end{array}\]

Here, \(\begin{align}({x_1},{y_1}) = \left( {\frac{1}{{\sqrt 3 }},\frac{{\sqrt 2 }}{{\sqrt 3 }}} \right)\end{align}\) and \(a = 1\) . Thus, the required equation is \(x + \sqrt 2 y = \sqrt 3 \)

**(b)** The equation of the tangent to \({x^2} + {y^2} + 2gx + 2fy + c = 0\) at \(({x_1},{y_1})\) is

\[\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&T({x_1},{y_1}) = 0\\\\ \Rightarrow & x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0 \end{array}\]

Here, \(({x_1},{y_1})\) is \(\begin{align}\left( {\frac{1}{2},2 + \frac{{\sqrt 3 }}{2}} \right)\end{align}\) and thus, the required equation is

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{x}{2} + \left( {2 + \frac{{\sqrt 3 }}{2}} \right)y - 1\left( {x + \frac{1}{2}} \right) - 2\left( {y + 2 + \frac{{\sqrt 3 }}{2}} \right) + 4 = 0\\\\\Rightarrow \qquad & \frac{{ - x}}{2} + \frac{{\sqrt 3 y}}{2} - \left( {\sqrt 3 + \frac{1}{2}} \right) = 0\\\\ \Rightarrow \qquad & - x + \sqrt 3 y - (2\sqrt 3 + 1) = 0\end{align}\]