Question

The number of integral values of $\lambda$ for which $x^2+y^2+\lambda x+(1-\lambda )y+5=0$ 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,
$$x^2+y^2+\lambda x+(1-\lambda )y+5=0$$
We compare the coefficients of $x$, coefficients of $y$ and the constant term of the equation with equation of general circle to have
$$g={\displaystyle \frac{-\lambda }{2}},f={\displaystyle \frac{-(1-\lambda )}{2}},c=5$$
So the radius of the circle is given by;
$$\begin{array}{rl}& r=\sqrt{{\left({\displaystyle \frac{-\lambda }{2}}\right)}^2+{(-{\displaystyle \frac{(1-\lambda )}{2}})}^2-5}\\ & \Rightarrow r=\sqrt{{\displaystyle \frac{{\lambda }^2}{4}}+{\displaystyle \frac{{(1-\lambda )}^2}{4}}-5}\end{array}$$
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;
$$\begin{array}{rl}& r\le 5\\ & \Rightarrow \sqrt{{\displaystyle \frac{{\lambda }^2}{4}}+{\displaystyle \frac{{(1-\lambda )}^2}{4}}-5}\le 5\end{array}$$
We square both sides to have;
$${\displaystyle \frac{{\lambda }^2}{4}}+{\displaystyle \frac{{(1-\lambda )}^2}{4}}-5\le 25$$
We multiply both sides by 4 to have;
$$\begin{array}{rl}& {\displaystyle \frac{{\lambda }^2}{4}}+{\displaystyle \frac{{(1-\lambda )}^2}{4}}-5\le 25\\ & \Rightarrow {\lambda }^2+{(1-\lambda )}^2-20\le 100\\ & \Rightarrow {\lambda }^2+{\lambda }^2+1-2\lambda -120\le 0\\ & \Rightarrow 2{\lambda }^2-2\lambda -119\le 0.......\left(1\right)\end{array}$$
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;
$$\begin{array}{rl}& \lambda ={\displaystyle \frac{2\pm \sqrt{{(-2)}^2-4\left(2\right)\left(119\right)}}{2\times 2}}\\ & \Rightarrow \lambda ={\displaystyle \frac{2\pm 2\sqrt{1-(-238)}}{2\times 2}}\\ & \Rightarrow \lambda ={\displaystyle \frac{1\pm \sqrt{239}}{2}}\end{array}$$
So the roots of the equations are
$$\begin{array}{rl}& \lambda ={\displaystyle \frac{1-\sqrt{239}}{2}}={\displaystyle \frac{1-15.4}{2}}=-7.2\left(\text{approximately}\right)\\ & \lambda ={\displaystyle \frac{1+\sqrt{239}}{2}}={\displaystyle \frac{1+15.4}{2}}=8.2\left(\text{approximately}\right)\end{array}$$
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 $(-g,-f)=({\displaystyle \frac{\lambda }{2}},{\displaystyle \frac{1-\lambda }{2}})$.