Circles
9.0 Tangents from a point to the circle
9.0 Tangents from a point to the circle
From a given point two tangents can be drawn to a circle which are real, coincident or imaginary according as the given point lies outside, on or inside the circle.
Proof: If the equation of circle is ${x^2} + {y^2} = {a^2}$ and tangent to the circle is $y = mx + a\sqrt {1 + {m^2}} $.
If any outside point is $P({x_1},{y_1})$ then, ${y_1} = m{x_1} + a\sqrt {1 + {m^2}} $
then, $${\left( {{y_1} - m{x_1}} \right)^2} = {a^2}\left( {1 + {m^2}} \right)$$ $${y_1}^2 + {m^2}{x_1}^2 - 2m{x_1}{y_1} = {a^2} + {a^2}{m^2}$$ $${m^2}\left( {{x_1}^2 - {a^2}} \right) - 2m{x_1}{y_1} + {y_1}^2 - {a^2} = 0...(1)$$
which is quadratic in $m$ which gives two values of $m$, real, coincident or imaginary, corresponding to any value of ${x_1}$ and ${y_1}$.
The tangents are real, coincident or imaginary according as the values of $m$ obtained from $(1)$ are real, coincident or imaginary.
Or, Discriminant$ > , = ,or < 0$
$4{x_1}^2{y_1}^2 - 4\left( {{x_1}^2 - {a^2}} \right)\left( {{y_1}^2 - {a^2}} \right) > , = ,or < 0$
and
${x_1}^2 + {y_1}^2 - {a^2} > , = ,or < 0$
i.e., $P\left( {{x_1},{y_1}} \right)$ lies outside, on or inside the circle ${x^2} + {y^2} = {a^2}$.
If $P$ lies outside the circle, then substituting these values of in equation of tangent $$y = mx + a\sqrt {1 + {m^2}} $$
we get the equation of tangents.
Question 20. Find the locus of point of intersection of tangents to the circle ${x^2} + {y^2} = {a^2}$ such that
(a) Angle between tangents is ${90^ \circ }$.
(b) Angle between tangents is ${45^ \circ }$.
Solution: Let $P(h,k)$ be the point of intersection of tangents to the circle ${x^2} + {y^2} = {a^2}$ as shown in figure $30$.
The equation of tangent of slope $m$ and passing through point $P(h,k)$ is $$y - k = m\left( {x - h} \right)$$ $$y - mx + hm - k = 0$$
And length of perpendicular drawn from centre $O(0,0)$ of circle is equal to the radius $a$ i.e., $a = \left| {\frac{{hm - k}}{{\sqrt {1 + {m^2}} }}} \right|$
Squaring both sides we get, $${a^2} + {a^2}{m^2} = {h^2}{m^2} + {k^2} - 2hkm$$ $${m^2}\left( {{h^2} - {a^2}} \right) - m\left( {2hk} \right) + \left( {{k^2} - {a^2}} \right) = 0$$
Sum of slopes ${m_1} + {m_2} = \frac{{2hk}}{{{h^2} - {a^2}}}...(1)$
Product of slopes ${m_1} \times {m_2} = \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}}...(2)$
(a) When angle between tangents is ${90^ \circ }$, Product of slopes ${m_1} \times {m_2} = \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}} = - 1$
or, ${k^2} - {a^2} = {a^2} - {h^2}$
${h^2} + {k^2} = 2{a^2}$
Therefore, the locus is ${x^2} + {y^2} = 2{a^2}$ which is the equation of director circle.
(b) When angle between tangents is ${45^ \circ }$, $\tan 45^\circ = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
$1 = \left| {\frac{{\surd {{({m_1} + {m_2})}^2} - 4{m_1}{m_2}}}{{1 + {m_1}{m_2}}}} \right|$
Put the values from equation $(1)$ and $(2)$ and squaring both sides we get, $${\left( {1 + \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}}} \right)^2} = \frac{{4{h^2}{k^2}}}{{{{\left( {{h^2} - {a^2}} \right)}^2}}} - 4 \times \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}}$$ $${\left( {\frac{{{h^2} - {a^2} + {k^2} - {a^2}}}{{{h^2} - {a^2}}}} \right)^2} = \frac{{4{h^2}{k^2}}}{{{{\left( {{h^2} - {a^2}} \right)}^2}}} - 4 \times \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}}$$ $${\left( {\frac{{{h^2} + {k^2} - 2{a^2}}}{{{h^2} - {a^2}}}} \right)^2} = \frac{{4{a^2}\left( {{h^2} + {k^2} - {a^2}} \right)}}{{{{\left( {{h^2} - {a^2}} \right)}^2}}}$$
On solving this we get, ${h^2} + {k^2} = \frac{{2{a^2}\left( {1 - 2{a^2}} \right)}}{{1 - 4{a^2}}}$
Therefore, the locus is $${x^2} + {y^2} = \frac{{2{a^2}\left( {1 - 2{a^2}} \right)}}{{1 - 4{a^2}}}$$
