Circles
10.0 Length of tangent from a point to a circle
10.0 Length of tangent from a point to a circle
Length of tangent from point $P({x_1},{y_1})$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is $\sqrt {{S_1}} $ where ${S_1} = {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0$
Proof: Let $PT$ and $PT'$ be two tangents from the given point $P({x_1},{y_1})$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$.
Then coordinates of centre of circle is $C(-g,-f)$ and radius $r = \sqrt {{g^2} + {f^2} - c} $.
Therefore, $CT = CT' = radius$.
In $\Delta PCT,PT = \sqrt {(P{C^2} - C{T^2})} $
$$PT = \sqrt {{{\left( {{x_1} + g} \right)}^2} + {{\left( {{x_1} + f} \right)}^2} - {g^2} - {f^2} + c} $$ $$ = \sqrt {{x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c} = \sqrt {{S_1}} = PT'$$
NOTE: For ${S_1}$, write the equation of circle in standard form with coefficient of ${x^2}=$coefficient of ${y^2}=1$ and making RHS of circle equal to $0$, then replace $(x,y)$ by $({x_1},{y_1})$.
Question 21. Find the length of tangents drawn from the point $(3,-4)$ to the circle $2{x^2} + 2{y^2} - 7x - 9y - 13 = 0$.
Solution: The equation of the given circle is $2{x^2} + 2{y^2} - 7x - 9y - 13 = 0$
or, $${x^2} + {y^2} - \frac{7}{2}x - \frac{9}{2}y - \frac{{13}}{2} = 0$$
Let $$S \equiv {x^2} + {y^2} - \frac{7}{2}x - \frac{9}{2}y - \frac{{13}}{2} = 0$$ Therefore, $${S_1} = {3^2} + - {4^2} - \frac{7}{2} \times 3 - \frac{9}{2} \times - 4 - \frac{{13}}{2}$$ $$ = 25 - \frac{{21}}{2} + 18 - \frac{{13}}{2} = 26$$
The length of tangent$ = \sqrt {{S_1}} = \sqrt {26} $.
