Circles
18.0 Chord of contact
18.0 Chord of contact
From an external point $({x_1},{y_1})$, two tangents can be drawn to a given circle. The chord $AB$ joining points of contact of two tangents is called the chord of contact of tangents.
Equation of chord of contact $AB$ to the circle ${x^2} + {y^2} = {a^2}$ can be find out using $T=0$ i.e., $$x{x_1} + y{y_1} = {a^2}$$
Proof: Let $A(x',y')$ and $B(x'',y'')$ be points of contact of tangents drawn from $P({x_1},{y_1})$ to the circle ${x^2} + {y^2} = {a^2}$. Then, equation of tangents $PA$ and $PB$ are
$$xx' + yy' = {a^2}$$ and $$xx'' + yy'' = {a^2}$$
respectively. Since both tangents pass through $P({x_1},{y_1})$ then, $${x_1}x' + {y_1}y' = {a^2}$$ and $${x_1}x'' + {y_1}y'' = {a^2}$$
Therefore, points $A(x',y')$ and $B(x'',y'')$ lie on $x{x_1} + y{y_1} = {a^2}$ which is the equation of chord of contact.
NOTE: Similarly if the equation of circle is given in general form i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$ then apply $T=0$ to find the chord of contact which can be written as $$x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$$
Question 27. Show that the area of triangle formed by the tangents from the point $P(4,3)$ to the circle ${x^2} + {y^2} = 9$ and the segment joining their points of contact is $7\frac{{17}}{{25}}$.
Solution: As shown in figure $52$, $AB$ is the chord of contact and its equation can be find out using $T=0$ i.e., $4x+3y=9$.
Perpendicular distance from centre of circle $C(0,0)$ to the chord of contact $AB$ is $OR = \left| {\frac{{0 + 0 - 9}}{{\sqrt {16 + 9} }}} \right| = \frac{9}{5}$
Apply Pythagoras theorem in $\Delta OAR$, $A{R^2} = O{A^2} - O{R^2}$
$$AR = \sqrt {{3^2} - {{\frac{9}{5}}^2}} = \sqrt {9 - \frac{{81}}{{25}}} = \frac{{12}}{5}$$
Therefore, $AB = 2AR = \frac{{24}}{5}$
Now, Distance between points $O$ and $P$ using distance formulae is $OP = \sqrt {{4^2} + {3^2}} = 5$.
Therefore, $PR = OP - OR = 5 - \frac{9}{5} = \frac{{16}}{5}$
Now, Area of $\Delta PAB = \frac{1}{2} \times AB \times RP = \frac{1}{2} \times \frac{{24}}{5} \times \frac{{16}}{5} = \frac{{192}}{{25}} = 7_{25}^{17}$
