Question

A family of linear functions is given by $f\left(x\right)=1+c(x+3)$ where $c\in R$. If a member of this family meets a unit circle centred at origin in two coincidence points then 'c' can be equal to

• A. $-3/4$
• B. $-1$
• C. $3/4$
• D. $1$

Answer 1

The correct option is A $-3/4$

Given function,

$f\left(x\right)=1+c(x+3)$ .....( 1 ) where $c\in R$

We know that,

General equation of circle of radius $a$ meets at the origin is

$x^2+y^2=a^2$

But it is unit circle then $a=1$,

Then, equation of circle is

$x^2+y^2=1^2$

$x^2+y^2=1$

Let $f\left(x\right)=y$

By equation $\left(1\right)$ and we get,

$y=1+c(x+3)$

$y=1+cx+3$ ......( 2 )

Using distance formula, at the origin to a line

$d=\left|\frac{0-c\times 0-1-3c}{\sqrt{1^2+c^2}}\right|=1$

$\left|\frac{-1-3c}{\sqrt{1^2+c^2}}\right|=1$

Taking square both side and we get,

${\left(\frac{1+3c}{\sqrt{1+c^2}}\right)}^2=1$

${(1+3c)}^2=1+c^2$

$1^2+{\left(3c\right)}^2+6c=1+c^2$

$1+9c^2+6c=1+c^2$

$9c^2+6c-c^2=0$

$8c^2+6c=0$

$2c(4c+3)=0$

$2c=0,4c+3=0$

$c=0,4c=-3$

$c=0,c=-\frac{3}{4}$

Hence, It is required solution.