Question

If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ cannot be equal to

• A. $\frac{1}{2}$
• B. $8$
• C. $50$
• D. $72$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{1}{2}$
B $8$
C $50$
D $72$

Given circle is $x^2+y^2-4x+2y+1=0$
$\Rightarrow (x-2{)}^2+(y+1{)}^2=2^2$
Thus any point parametric point on this circle is given by, $(2+2\cos \theta ,-1+2\sin \theta )$
Now Let $z=x^2+y^2-10x-6y+34=(x-5{)}^2+(y-3{)}^2=(2\cos \theta -3{)}^2+(2\sin \theta -4{)}^2$
$\Rightarrow z=4{\cos }^2\theta +9-12\cos \theta +4{\sin }^2\theta -16\sin \theta +16=29-4(3\cos \theta +4\sin \theta )$
We know $-5\le 3\cos \theta +4\sin \theta \le 5$
Hence $z\in [9,49]$