# The set of all real values of $\lambda$ for which exactly two common tangents can be drawn to the circles ${x^2} + {y^2} - 4x - 4y + 6 = 0$ and ${x^2} + {y^2} - 10x - 10y + \lambda = 0$ is the interval:A. $(18,42)$B. $(12,32)$C. $(12,24)$D. $(18,48)$

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Hint: Using the general equation of a circle we compare the values of the two circles and write the centers and radius of the circles. Then using the fact that the circles can have two common tangents only if they are intersecting circles and the tangent for the first circle is tangent for the second circle as well. Using the values of distance between centers and radii of two circles we will relate when the two circles will be intersecting and substituting the values we will solve for $\lambda$.
* General equation of a circle is given by ${x^2} + {y^2} + 2gx + 2fy + c = 0$ where the center of circle is given by $( - g, - f)$ and the radius of circle is given by $\sqrt {{g^2} + {f^2} - c}$

Let us assume the two circles as ${C_1}$ and ${C_2}$.
Let equation of ${C_1}$ be ${x^2} + {y^2} - 4x - 4y + 6 = 0$
Let equation of ${C_2}$ be ${x^2} + {y^2} - 10x - 10y + \lambda = 0$
We know that general equation of circle is given by ${x^2} + {y^2} + 2gx + 2fy + c = 0$
On comparing the equation for ${C_1}$, we get
$\Rightarrow 2g = - 4,2f = - 4 \\ \Rightarrow g = - 2,f = - 2 \\$
SO, the center of circle ${C_1}$ is $( - g, - f) = ( - ( - 2), - ( - 2)) = (2,2)$
And radius of circle ${C_1}$ is $\sqrt {{g^2} + {f^2} - c} = \sqrt {{{( - 2)}^2} + {{( - 2)}^2} - 6}$
$\Rightarrow \sqrt {{g^2} + {f^2} - c} = \sqrt {4 + 4 - 6} \\ \Rightarrow \sqrt {{g^2} + {f^2} - c} = \sqrt 2 \\$
So, the radius of the circle ${C_1}$ is $\sqrt 2$.
On comparing the equation for ${C_2}$, we get
$\Rightarrow 2g = - 10,2f = - 10 \\ \Rightarrow g = - 5,f = - 5 \\$
SO, the center of circle ${C_2}$ is $( - g, - f) = ( - ( - 5), - ( - 5)) = (5,5)$
And radius of circle ${C_2}$ is $\sqrt {{g^2} + {f^2} - c} = \sqrt {{{( - 5)}^2} + {{( - 5)}^2} - \lambda }$
$\Rightarrow \sqrt {{g^2} + {f^2} - c} = \sqrt {25 + 25 - \lambda } \\ \Rightarrow \sqrt {{g^2} + {f^2} - c} = \sqrt {50 - \lambda } \\$
So, the radius of the circle ${C_2}$ is $\sqrt {50 - \lambda }$.
Now we know that there can be exactly two tangents that are common to two circles, then the two circles are intersecting circles.

If the sum of radii is equal to distance between centers then the circles will just touch and then we can form more than two common tangents as the third tangent will be touching the point where the circles meet.
If the sum of radii is less than distance between centers then the circles will be far apart then also we can form more than two common tangents as the tangents crossing the midpoint of the line joining two centers will touch both circles.
For the circles to be intersecting, the sum of their radii should always be greater than the distance between their centers
i.e. ${r_1} + {r_2} > d({C_1}{C_2})$ … (1)
We know distance between any two points $(a,b),(c,d)$ is given by $\sqrt {{{(a - c)}^2} + {{(b - d)}^2}}$
Here the two points are $(2,2),(5,5)$
$\Rightarrow d({C_1}{C_2}) = \sqrt {{{(2 - 5)}^2} + {{(2 - 5)}^2}} \\ \Rightarrow d({C_1}{C_2}) = \sqrt {{{( - 3)}^2} + {{( - 3)}^2}} \\ \Rightarrow d({C_1}{C_2}) = \sqrt {{3^2} + {3^2}} \\ \Rightarrow d({C_1}{C_2}) = \sqrt {2 \times {3^2}} \\ \Rightarrow d({C_1}{C_2}) = 3\sqrt 2 \\$
Also, ${r_1} = \sqrt 2 ,{r_2} = \sqrt {50 - \lambda }$
Substituting the value of distance between the centers and the radius of two circles in equation (1) we get
$\Rightarrow \sqrt 2 + \sqrt {50 - \lambda } > 3\sqrt 2$
Shifting the constants to one side of the inequality
$\Rightarrow \sqrt {50 - \lambda } > 3\sqrt 2 - \sqrt 2 \\ \Rightarrow \sqrt {50 - \lambda } > 2\sqrt 2 \\$
Squaring both sides of the inequality
$\Rightarrow {\left( {\sqrt {50 - \lambda } } \right)^2} > {\left( {2\sqrt 2 } \right)^2} \\ \Rightarrow 50 - \lambda > 8 \\$
Shifting the constants to one side of the inequality
$\Rightarrow - \lambda > 8 - 50 \\ \Rightarrow - \lambda > - 42 \\$
Multiplying both sides by -1
$\Rightarrow - \lambda \times - 1 > - 42 \times - 1 \\ \Rightarrow \lambda < 42 \\$ {sign changes in inequality when multiplied by negative sign}
The upper bound of the set is $42$
Also, the condition to make two circles not concentric is that the difference of their radii should always be less than or equal to the distance between the centers of two circles.
i.e. $\left| {{r_1} - {r_2}} \right| < d({C_1}{C_2})$
Substituting the value of distance between the centers and the radius of two circles in equation (1) we get
$\Rightarrow \sqrt {50 - \lambda } - \sqrt 2 < 3\sqrt 2$
Shifting the constants to one side of the inequality
$\Rightarrow \sqrt {50 - \lambda } < 3\sqrt 2 + \sqrt 2 \\ \Rightarrow \sqrt {50 - \lambda } < 4\sqrt 2 \\ \Rightarrow \sqrt {50 - \lambda } < 4\sqrt 2 \\$
Squaring both sides of the inequality
$\Rightarrow {\left( {\sqrt {50 - \lambda } } \right)^2} < {\left( {4\sqrt 2 } \right)^2} \\ \Rightarrow 50 - \lambda < 32 \\$
Shifting the constants to one side of the inequality
$\Rightarrow - \lambda < 32 - 50 \\ \Rightarrow - \lambda < - 18 \\$
Multiplying both sides by -1
$\Rightarrow - \lambda \times - 1 < - 18 \times - 1 \\ \Rightarrow \lambda > 18 \\$ {sign changes in inequality when multiplied by negative sign}
The lower bound of the set is $18$ So, the set of real values of $\lambda$ is $(18,42)$.

So, the correct answer is “Option A”.

Note: Students are likely to make mistakes while doing calculation with inequalities, always keep in mind that the sign of the inequality changes when we multiply it with a negative number.