Question

Find the least value of $x$ satisfying the relation $x^2+y^2=6x-8y$ .

• A. $-2$
• B. $9$
• C. $-7$
• D. $-3$

Answer 1

The correct option is A $-2$

The given curve is, $x^2+y^2=6x-8y$

$\Rightarrow (x^2-6x)+(y^2+8y)=0$

$\Rightarrow (x^2-6x+9)+(y^2+8y+16)=9+16$, make perfect square

$\Rightarrow (x-3{)}^2+(y+4{)}^2=25=5^2$

Clearly above equation represents circle having centre $(3,4)$ and radius $5$

So any parametric point the above circle is taken as

$x-3=5\cos \theta$ and $y+4=5\cos \theta$

$\Rightarrow x=3+5\cos \theta$

Now we know that $-1\le \cos \theta \le 1$

Hence $x_{max}=3+5\cdot (-1)=3-5=-2$