Circles
20.0 Director Circle
20.0 Director Circle
The locus of the point of intersection of two perpendicular tangents to a given circle is known as its director circle.
The equation of director circle of the circle ${x^2} + {y^2} = 2{a^2}$ i.e., the radius of director circle is $\sqrt 2 a$ with the same centre as that of given circle.
Question 28. Two circles are given, ${S_1} \equiv {x^2} + {y^2} + 2x + 2ky + 6 = 0$ and ${S_2} \equiv {x^2} + {y^2} + 2ky + k = 0$. Find value of $k$ if the circles are orthogonal.
Solution: From the given two equations of circle, ${g_1} = 1$, ${f_1} = k$, ${c_1} = 6$, ${g_2} = 0$, ${f_2} = k$, ${c_2} = k$
If the circles are orthogonal, they satisfy the condition ${c_1} + {c_2} = 2{g_1}{g_2} + 2{f_1}{f_2}$.
Therefore, $$6 + k = 2 \times 1 \times 0 + 2 \times k \times k$$ $$2{k^2} - k - 6 = 0$$ $$2{k^2} - 4k + 3k - 6 = 0$$ $$\left( {k - 2} \right)\left( {2k + 3} \right) = 0$$
Therefore, $$k = 2{\text{ or }}k = \frac{{ - 3}}{2}$$
