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 coincident point then '$c$' be equal to

• A. $\frac{-3}{4}$
• B. $0$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

The correct option is A $\frac{-3}{4}$

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

Circle : $x$ ${}^2$ + y${}^2$ = 1

$\Rightarrow$ $x$ ${}^2$ + [(1+c($x$ + 3))${}^2$ } = 1

$\Rightarrow$ $x$ ${}^2$ + [1+c($x$ + 3)${}^2$ +2c($x$ +3)} = 1

$\Rightarrow$ $x$ ${}^2$ + $c$ ${}^2$ ($x$ + 3)${}^2$ +2c($x$ +3)} = 0.

$\Rightarrow$ $x$ ${}^2$ + 2c ($x$ + 3)+ $c$ ${}^2$ ($x$ ${}^2$ + 9 + 6$x$) = 0.

$\Rightarrow$ $x$ ${}^2$ + 2c$x$ +6c + $c$ ${}^2$ $x$ ${}^2$ +9c${}^2$ + 6c${}^2$ $x$ = 0

$\Rightarrow$ $x$ ${}^2$ ( $c$ ${}^2$ + 1) + $x$ (2c + 6c${}^2$ ) + (6c+9c${}^2$) +0.

Discriminant = 0 ( $\therefore$ coincident points).

$\Rightarrow$ (2c +6c${}^2$)${}^2$ = 4( c${}^2$+ 1) (6c + 9c${}^2$)

$\Rightarrow$ 4c${}^2$ +36c${}^4$ + 24c${}^3$ = (4c${}^2$ + 4) (6c + 9c${}^2$ )

$\Rightarrow$ 4c${}^2$ +36c${}^4$ + 24c${}^3$ = (4c${}^2$ + 4) (6c + 9c${}^2$ )

$\Rightarrow$ 4c${}^2$ +36c${}^4$ + 24c${}^3$ = 24c${}^3$ +36c${}^4$ + 24c + 36c${}^2$

$\Rightarrow$ 4c${}^2$ - 36c${}^2$ = 24c

$\Rightarrow$ 32c${}^2$ = 24c

$\Rightarrow$ -4c = 3 $\Rightarrow$ c = $\frac{-3}{4}$.