Question

The number of integral values of $\lambda $ for which ${{x}^{2}}+{{y}^{2}}+\lambda x+\left( 1-\lambda \right)y+5=0$ is the equation of a circle whose radius cannot exceed 5, is
A. 14
B. 18
C. 16
D. None

Answer 1

Hint: We compare the given equation of circle with general equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ and find the radius of the circle as $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. We use the wavy curve method for what values of $\lambda $ the radius is less than 5.

Complete step-by-step solution:
We know that the general equation of circle in two variables is given by the equation,
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]
We know the radius $r$ of the above circle is given by
\[r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]
We are given the equation from the question with parameter $\lambda $ as,
\[{{x}^{2}}+{{y}^{2}}+\lambda x+\left( 1-\lambda \right)y+5=0\]
We compare the coefficients of $x$, coefficients of $y$ and the constant term of the equation with equation of general circle to have
\[g=\dfrac{-\lambda }{2},f=\dfrac{-\left( 1-\lambda \right)}{2},c=5\]
So the radius of the circle is given by;
\[\begin{align}
  & r=\sqrt{{{\left( \dfrac{-\lambda }{2} \right)}^{2}}+{{\left( -\dfrac{\left( 1-\lambda \right)}{2} \right)}^{2}}-5} \\
 & \Rightarrow r=\sqrt{\dfrac{{{\lambda }^{2}}}{4}+\dfrac{{{\left( 1-\lambda \right)}^{2}}}{4}-5} \\
\end{align}\]
We are given in the question that the radius of the given circle has to be less than or equal to 5. So we have;
\[\begin{align}
  & r\le 5 \\
 & \Rightarrow \sqrt{\dfrac{{{\lambda }^{2}}}{4}+\dfrac{{{\left( 1-\lambda \right)}^{2}}}{4}-5}\le 5 \\
\end{align}\]
We square both sides to have;
\[\dfrac{{{\lambda }^{2}}}{4}+\dfrac{{{\left( 1-\lambda \right)}^{2}}}{4}-5\le 25\]
We multiply both sides by 4 to have;
\[\begin{align}
  & \dfrac{{{\lambda }^{2}}}{4}+\dfrac{{{\left( 1-\lambda \right)}^{2}}}{4}-5\le 25 \\
 & \Rightarrow {{\lambda }^{2}}+{{\left( 1-\lambda \right)}^{2}}-20\le 100 \\
 & \Rightarrow {{\lambda }^{2}}+{{\lambda }^{2}}+1-2\lambda -120\le 0 \\
 & \Rightarrow 2{{\lambda }^{2}}-2\lambda -119\le 0.......\left( 1 \right) \\
\end{align}\]
We see that in the left hand side of the equation there is a quadratic equation in $\lambda $. Let us find the zeroes of the quadratic equation $2{{\lambda }^{2}}-2\lambda -119=0$ using the quadratic formula. We have;
\[\begin{align}
  & \lambda =\dfrac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 2 \right)\left( 119 \right)}}{2\times 2} \\
 & \Rightarrow \lambda =\dfrac{2\pm 2\sqrt{1-\left( -238 \right)}}{2\times 2} \\
 & \Rightarrow \lambda =\dfrac{1\pm \sqrt{239}}{2} \\
\end{align}\]
So the roots of the equations are
\[\begin{align}
  & \lambda =\dfrac{1-\sqrt{239}}{2}=\dfrac{1-15.4}{2}=-7.2\left( \text{approximately} \right) \\
 & \lambda =\dfrac{1+\sqrt{239}}{2}=\dfrac{1+15.4}{2}=8.2\left( \text{approximately} \right) \\
\end{align}\]
Since (1) is an inequality let us check using a wavy curve method and find for what values of $\lambda $ the inequality (1) satisfies. We have;

So $\lambda $ has to lie in between $-7.2$ to 8.2 to satisfy inequality (1). So the possible integral values of $\lambda $ are $-7,-8,...,8$. So there are a total 16 integral values of $\lambda $. So the correct choice is C. \[\]

Note: We note that by putting $\lambda =0$ in inequality (1) to quickly find the positive negative signs for the wavy curve. We should try to estimate the square root $\sqrt{239}$ using nearest perfect square ${{15}^{2}}=225,{{16}^{2}}=256$ to find the integral values quickly. We can also find the centre of the circle as $\left( -g,-f \right)=\left( \dfrac{\lambda }{2},\dfrac{1-\lambda }{2} \right)$.