13.0 Pair of tangents

  Circles

The combined equation of pair of tangents drawn from a point $P({x_1},{y_1})$ to a circle ${x^2} + {y^2} = {a^2}$ can be written as $$S{S_1} = {T^2}$$ i.e., $$\left( {{x^2} + {y^2} - {a^2}} \right)\left( {{x_1}^2 + {y_1}^2 - {a^2}} \right) = {\left( {x{x_1} + y{y_1} - {a^2}} \right)^2}$$

Question 23. Find the equation of pair of tangents drawn to the circle ${x^2} + {y^2} - 2x + 4y = 0$ from the point $(0,1)$.

Solution: The equation of given circle is $S \equiv {x^2} + {y^2} - 2x + 4y = 0$ with centre of circle $(1,-2)$ and radius $r = \sqrt {{g^2} + {f^2} - c} = \sqrt {{1^2} + - {2^2} - 0} = \sqrt 5 $

And,

$P(0,1)$ is the point from which the pair of tangents is drawn. For point $P$, $${S_1} = {0^2} + {1^2} - 2 \times 0 + 4 \times 1 = 5$$

Since value of ${S_1}$ is greater than radius of circle, point lies outside the circle. $$T \equiv x \times 0 + y \times 1 - \left( {x + 0} \right) + 2\left( {y + 1} \right)$$ $$T \equiv - x + 3y + 2$$

Now, the equation of pair of tangents drawn to the circle ${x^2} + {y^2} - 2x + 4y = 0$ from the point $(0,1)$ can be written as $S{S_1} = {T^2}$

$$5\left( {{x^2} + {y^2} - 2x + 4y} \right) = {( - x + 3y + 2)^2}$$ $$5{x^2} + 5{y^2} - 10x + 20y = {x^2} + 9{y^2} + 4 - 6xy - 4x + 12y$$ $$4{x^2} - 4{y^2} - 6x + 8y + 6xy - 4 = 0$$ $$2{x^2} - 2{y^2} - 3x + 4y + 3xy - 2 = 0$$

NOTE: To find the separate equations of tangents, make the above equation as a quadratic equation in $x$ and then find the value of $x$ in terms of $y$ to find the separate equation of tangents.