Question

If the equation $x^2+y^2-10x+21=0$ has real roots $x=\alpha$ and $y=\beta$ then

• A. $3\le x\le 7$
• B. $3\le y\le 7$
• C. $-2\le y\le 2$
• D. $-2\le x\le 2$

Answer 1

Answer: (A), (C)

The correct options are
A $3\le x\le 7$
C $-2\le y\le 2$

For the given equation $x^2+y^2-10x+21=0$, we can write the equation as
$x^2-10x+(21+y^2)=0$
Above equation is now quadratic in $x$ and the equation has real roots
$\Rightarrow D\ge 0$
$\Rightarrow 100-4(21+y^2)\ge 0$
$\Rightarrow y^2\le 4$
$\Rightarrow -2\le y\le 2$
Also, the above equation can be written as:
$y^2=-x^2+10x-21$
Here, the equation will have real roots
$\Rightarrow -x^2+10x-21\ge 0$
$\Rightarrow (x-7)(x-3)\le 0$
$3\le x\le 7$

Alternate solution:
Given equation:
$x^2+y^2-10x+21=0\Rightarrow (x-5{)}^2+y^2=4$


So the roots of the equation will lie on the circle,
$3\le x\le 7-2\le y\le 2$