Examples On Tangents To Circles Set-1

Example - 17

What is the length of the tangent to \(S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0\) drawn from an external point \(P({x_1},{y_1})\) ?

Solution:                                                                  

The length \(PA\) (or \(PB\)) can be evaluated by a simple application of the Pythagoras theorem.

In \(\Delta PAO,\) observe that

\[\begin{align}  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P{A^2} =& P{O^2} - A{O^2} \hfill \\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =& \left\{ {{{({x_1} + g)}^2} + {{({y_1} + f)}^2}} \right\} - {\left\{ {\sqrt {{g^2} + {f^2} - c} } \right\}^2} \hfill \\\\  &\left\{ {\because AO{\text{ is the radius}}} \right\} \hfill \\\\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =& x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right) \hfill \\ \end{align} \]

The equation of the circle being represented by \(S = 0,\)  we can denote the RHS obtained in (1) by \(S({x_1},{y_1}).\) Thus, the length of the tangent can be written concisely as

\[\fbox{$\begin{array}{*{20}{c}} {PA = \sqrt {S({x_1},{y_1})} }\end{array}$}\]

For example, the length of the tangent from \((4, 4)\) to \({x^2} + {y^2} - 2x - 4y + 4 = 0\) will be

\[\begin{array}{l}l = \sqrt {{4^2} + {4^2} - 2 \times 4 - 4 \times 4 + 4} \\\,\,\, = \sqrt {12} \\\,\,\, = 2\sqrt 3 \end{array}\]

In the next example, we discuss how to write the equation to the pair of tangents \(PA\) and \(PB\).

Example - 18

From an external point \(P({x_1},{y_1}),\) (two) tangents are drawn to the circle \(S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0\) . These tangents touch the circle at \(A\) and \(B\). Find the joint equation of \(PA\) and \(PB\).

Solution: Consider any point  \((h,k)\) lying on the tangents drawn from \(P\) to \(S\).

Since we know two points on the line \(PA\), we can use the two-point form to write its equation:

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{{y - k}}{{x - h}} = \frac{{{y_1} - k}}{{{x_1} - h}}\\\\ \Rightarrow \qquad & x({y_1} - k) - y({x_1} - h) = h({y_1} - k) - k({x_1} - h)\\\\ \Rightarrow \qquad & x({y_1} - k) - y({x_1} - h) + (k{x_1} - h{y_1}) = 0\end{align}\]

Since \(PA\) is a tangent to \(S\), its distance from the centre of the circle, \(( - g, - f),\) must equal the radius.

This gives

\[\frac{{{{( - g({y_1} - k) + f({x_1} - h) + k{x_1} - h{y_1})}^2}}}{{{{({x_1} - h)}^2} + {{({y_1} - k)}^2}}} = {g^2} + {f^2} - c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

To write the equation of the pair of lines in conventional form, we use \((x,y)\) instead of \((h,k)\) in (1), above. Subsequent (lengthy!) rearrangements give:

\[{\left\{ {x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c} \right\}^2} = \left( {{x^2} + {y^2} + 2gx + 2fy + c} \right)\left( {x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c} \right)\] The left hand side can be written concisely as \({(T({x_1},{y_1}))^2}\) as described earlier whereas the right hand side can be written concisely as \(S(x,y)S({x_1},{y_1}).\) Thus, the equation to the pair of tangents can be written concisely as

\[\fbox{$\begin{array}{*{20}{c}} {{T^2}({x_1},{y_1}) = S(x,y)S({x_1},{y_1})}\end{array}$}\]

This relation be written in an even shorter form as simply \({T^2} = S{S_1}\) .

Example - 19

Find the equation to the pair of tangents drawn from the origin to the circle \({x^2} + {y^2} - 4x - 4y + 7 = 0\)  .

Solution: We use the relation obtained in the last example, \({T^2} = S{S_1},\) to write the desired equation. Here, \(({x_1},{y_1})\) is (0, 0) while  \(g =  - 2,f =  - 2\) and \(c = 7\) . Thus the joint equation is

\[\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{T^2}(0,0) = S(x,y)S(0,0)\\\\ \Rightarrow  & {( - 2x - 2y + 7)^2} = ({x^2} + {y^2} - 4x - 4y + 7)(7)\\\\ \Rightarrow  & 4{x^2} + 4{y^2} + 49 + 8xy - 28x - 28y = 7{x^2} + 7{y^2} - 28x - 28y + 49\\\\ \Rightarrow  & 3{x^2} - 8xy + 3{y^2} = 0\end{array}\]

As expected, since the tangents have been drawn from the origin, the obtained equation is a homogenous one.