Question

The two circles ${x^2} + {y^2} = ax$ and ${x^2} + {y^2} = {c^2}(c > 0)$ touch each other if:
A. $2\left| a \right| = c$
B. $\left| a \right| = c$
C. $a = 2c$
D. $\left| a \right| = 2c$

Answer 1

Hint: In the above question, we are given two circles. We will firstly try to find the radius and centre of both the circles. And we have to find the condition if both circles touch each other. Then we will directly use the condition for the circles touching each other.Now we know that two circles touch each other if $\left| {{C_1}{C_2}} \right| = {r_1} \pm {r_2}$ where ${C_1},{C_2}$ are the centres of the two circles and ${r_1},{r_2}$ be the radius of the two circles.Using this concept we try to solve the question.

Complete step-by-step answer:
We are given equation of the two circles:
${x^2} + {y^2} = ax$$ - - - - (1)$
${x^2} + {y^2} = {c^2}(c > 0)$$ - - - - (2)$
Now we know the general equation of the circle with centre $C(a,b)$ and radius $r$ is
${(x - a)^2} + {(y - b)^2} = {r^2}$$ - - - - - (3)$
Now we know that centre of the circle is $\left( {\dfrac{{ - coefficient{\text{ }}ofx}}{2},\dfrac{{ - coefficient{\text{ }}ofy}}{2}} \right)$
And radius of the circle $ = {\left( {{{\left( {\dfrac{{{\text{coefficient of x}}}}{2}} \right)}^2} + {{\left( {\dfrac{{{\text{coefficient of y}}}}{2}} \right)}^2} - {\text{constant term}}} \right)^{\dfrac{1}{2}}}$
Let ${C_1},{C_2}$be the centres of the two circles (1) and (2).
Now we will consider (1), and find the radius and centre of the circle using (3), the general equation of the circle, we get
${x^2} + {y^2} = ax$
${x^2} - ax + {y^2} = 0$$ - - - (4)$
Now we know that centre of the circle is $\left( {\dfrac{{ - coefficient{\text{ }}ofx}}{2},\dfrac{{ - coefficient{\text{ }}ofy}}{2}} \right)$
Which is ${C_1} = $$\left( {\dfrac{a}{2},0} \right)$
And radius of the circle $ = {\left( {{{\left( {\dfrac{{{\text{coefficient of x}}}}{2}} \right)}^2} + {{\left( {\dfrac{{{\text{coefficient of y}}}}{2}} \right)}^2} - {\text{constant term}}} \right)^{\dfrac{1}{2}}}$
So, using (4), we get,
${r_1} = \sqrt {\dfrac{{{a^2}}}{4} + 0 - 0} $
${r_1} = \left| {\dfrac{a}{2}} \right|$
Hence, we got the centre point ${C_1} = $$\left( {\dfrac{a}{2},0} \right)$ and radius ${r_1} = \left| {\dfrac{a}{2}} \right|$ of the (1)
Now we will consider (1), and find the radius and centre of the circle using (3), the general equation of the circle, we get,
$
  {x^2} + {y^2} = {c^2}(c > 0) \\
  {\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {c^2} \\
 $
Here after comparing with (3), we get:
${r_2}$$ = c$
${C_2} = (0,0)$
Hence, we got the centre point ${C_2} = (0,0)$ and radius ${r_2}$$ = c$ of the (2)
Now, we have to find the condition if both circles touch each other.
So, we know that two circles touch each other if
$\left| {{C_1}{C_2}} \right| = {r_1} \pm {r_2}$ $ - - - - - (7)$
Where ${C_1},{C_2}$are the centres of the two circles and ${r_1},{r_2}$be the radius of the two circles.
Now using (7) for the given circles (1) and (2), where ${r_1}$$ = \left| {\dfrac{a}{2}} \right|$ and ${r_2}$$ = c$
${C_1}{C_2}$is the distance between the centres of the two circles, we get:
$\left| {{C_1}{C_2}} \right| = {r_1} \pm {r_2}$
Now we know that the distance formula for the two points $A = ({x_1},{y_1})$ and $B = ({x_2},{y_2})$ is
$AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Now using the same formula for the point ${C_1}{C_2}$, where ${x_1} = \dfrac{a}{2},{y_1} = 0{\text{ and }}{x_2} = 0,{y_2} = 0$, we get
$\left| {{C_1}{C_2}} \right| = \left| {\dfrac{a}{2}} \right| \pm c$
$\sqrt {{{(0 - \dfrac{a}{2})}^2} + {{(0 - 0)}^2}} $$ = \left| {\dfrac{a}{2}} \right| \pm c$
$\sqrt {{{\dfrac{a}{4}}^2}} $$ = \left| {\dfrac{a}{2}} \right| \pm c$
Squaring both the sides we get:
${\left( {\sqrt {{{\dfrac{a}{4}}^2}} } \right)^2} = {\left( {\left| {\dfrac{a}{2}} \right| \pm c} \right)^2}$
Now using the algebraic identity ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$, we get
$\dfrac{{{a^2}}}{4} = {\left| {\dfrac{a}{2}} \right|^2} + {c^2} \pm 2\left| {\dfrac{a}{2}} \right|c$
Now we know that ${\left| x \right|^2} = {x^2}$, so, we get,
${\dfrac{a}{4}^2}$$ = {\dfrac{a}{4}^2} + {c^2} \pm \left| a \right|c$
Now cancelling ${\dfrac{a}{4}^2}$ from both sides we get
$0 = {c^2} \pm \left| a \right|c$
${c^2} \pm \left| a \right|c = 0$
Now taking $c$ common from L.H.S., we get
$c(c \pm \left| a \right|) = 0$
$c = 0{\text{ or }}c \pm \left| a \right| = 0$
Now we are also given that:
$c > 0$, so, we cannot include $c = 0$, so we have,
$(c \pm \left| a \right|) = 0$
$\left| a \right| = \mp c$
$\left| a \right| = - c$ or $\left| a \right| = + c$
We know that $c > 0$
So $ - c < 0$
So, we can say that $ - c$ is a negative number
And $\left| a \right|$ is a positive number because modulus of any number is a positive number so we get that
$\left| a \right| \ne - c$ because a positive number is not equal to the negative number.
So, we will drop the equation $\left| a \right| = - c$
So, we get $\left| a \right| = c$
The two circles ${x^2} + {y^2} = ax$ and ${x^2} + {y^2} = {c^2}(c > 0)$ touch each other if $\left| a \right| = c$

So, the correct answer is “Option B”.

Note: In this question for the circle ${x^2} + {y^2} = ax$, we have taken its radius ${r_1} = \left| {\dfrac{a}{2}} \right|$ because radius is always positive and we have no idea about the sign of $ - a$. So, we have taken the modulus of $\dfrac{a}{2}$ as its radius. Without the modulus, it will be incorrect. Hence these things should be kept in mind while solving such problems.