Question

A circle touches a straight line $lx + my + n = 0$ & cuts the circle ${x^2} + {y^2} = 9$ orthogonally. The locus of centres of each circle is:
A. ${(lx + my + n)^2} = ({l^2} + {m^2})({x^2} + {y^2} - 9)$
B. ${(lx + my - n)^2} = ({l^2} + {m^2})({x^2} + {y^2} + 9)$
C. ${(lx + my + n)^2} = ({l^2} + {m^2})({x^2} + {y^2} + 9)$
D. ${(lx + my - n)^2} = ({l^2} + {m^2})({x^2} + {y^2} - 9)$

Answer 1

Hint: Here, we will use the concept of the tangent of the circle and the intersection of two circles orthogonally at right angles. Follow step by step with the help of the diagrams needed. Place given values and simplify accordingly.

Complete step-by-step answer:

Let us consider the centre of the circle is $(h,k)$ and as per the property the tangent touches the circle perpendicularly and the distance between the tangent and the circle is the radius.

Therefore, radius $r = \left| {\dfrac{{(lh + mk + n)}}{{\sqrt {{l^2} + {m^2}} }}} \right|$

Take square on both the side of the equation –

${r^2} = \dfrac{{{{(lh + mk + n)}^2}}}{{{{(\sqrt {{l^2} + {m^2}} )}^2}}}$

Square and square root cancel each other in the denominator on the right hand side of the equation.

${r^2} = \dfrac{{{{(lh + mk + n)}^2}}}{{{l^2} + {m^2}}}$ ..... (A)

Given that two circles intersect each other orthogonally means at right angles. Therefore, the given equation as centre $(0,0) \Rightarrow {r_1}^2 = 9$

Let us consider two circles ${C_1}{\text{ and }}{{\text{C}}_2}$ with the radius $r$ and ${r_1}$ which makes right angle at point A.

In $\Delta {C_1}A{C_2}$ , $\angle A$ is the right angle.

Therefore, by using Pythagoras theorem,

$A{C_1}^2 + A{C_2}^2 = {C_1}{C_2}^2$

Place the values in the above equation –

$\Rightarrow {r^2} + {r_1}^2 = {\left( {\sqrt {{h^2} + {k^2}} } \right)^2}$

Square and square root cancel each other on right hand side of the equation-

$\Rightarrow {r^2} + {r_1}^2 = {h^2} + {k^2}$

Place values by using the equation (A)

$\Rightarrow \dfrac{{{{(lh + mk + n)}^2}}}{{{l^2} + {m^2}}} + 9 = {h^2} + {k^2}$

When the term is moved from one side to another, the sign also changes from positive to negative and vice-versa.

$\Rightarrow \dfrac{{{{(lh + mk + n)}^2}}}{{{l^2} + {m^2}}} = {h^2} + {k^2} - 9$

When the term is in the division at one side, moved to the opposite side then it goes to the multiplicative with the terms on the opposite side.

$\Rightarrow {(lh + mk + n)^2} = \left( {{h^2} + {k^2} - 9} \right)\left( {{l^2} + {m^2}} \right)$

Replace $(h,k) = (x,y)$ for the co-ordinate

$\Rightarrow {(lx + my + n)^2} = \left( {{x^2} + {y^2} - 9} \right)\left( {{l^2} + {m^2}} \right)$

Re-arrange the terms as per the requirement –

$\Rightarrow {(lx + my + n)^2} = \left( {{l^2} + {m^2}} \right)\left( {{x^2} + {y^2} - 9} \right)$

So, the correct answer is “Option A”.

Note: Be careful while using the radius formula and simplification of the equations. Construct the required equation correctly using Pythagora's theorem, the answer solely depends on the conditions applied.