Examples On Tangents To Circles Set-1

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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.