Question

Any point on the circle \[{{x}^{2}}+{{y}^{2}}-4x-4y+4=0\] can be taken as
(a) \[\left( 2+2\cos \theta ,2+2\sin \theta \right)\]
(b) \[\left( 2-2\cos \theta ,2-2\sin \theta \right)\]
(c) \[\left( 2-2\cos \theta ,2+2\sin \theta \right)\]
(d) \[\left( 2+2\cos \theta ,2-2\sin \theta \right)\]

Answer 1

Hint: We need to first convert the circle into its center and radius form. And then we need to use the formula \[\left( x+r\cos \theta ,y+r\sin \theta \right)\].

Complete step-by-step answer:
From the question, we are given the circle: \[{{C}_{1}}={{x}^{2}}+{{y}^{2}}-4x-4y+4=0....\left( i \right)\]
We have to find the general point on a given circle. We know that, general equation of circle is:
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]
By comparing it with equation (i), we get
\[\begin{align}
  & 2g=-4,\text{ }2f=-4,c=4 \\
 & g=-2,\text{ }f=-2 \\
\end{align}\]
We know that, center of circle = (– g, – f) = (2, 2)
\[\text{Radius of circle }=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]
\[=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}-4}\]
\[=\sqrt{4}\]
\[=2\text{ units}\]
Circle with center (2, 2) and radius 2 units is as follows:

We know that, equation of circle with (h, k) center and radius r is:
\[={{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}....\left( ii \right)\]
Therefore, equation of above circle with center (2, 2) and radius 2 units is:
\[={{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( 2 \right)}^{2}}\]
General point on any circle is:
\[\left( x,y \right)=\left( h+r\cos \theta ,k+r\sin \theta \right)\]
Putting the required values of h = k = 2, r = 2, we get
\[\left( x,y \right)=\left( 2+2\cos \theta ,2+2\sin \theta \right)\]
Therefore, option (a) is the correct answer.

Note: Instead of taking into account the formula of center and radius of circle, we can directly rearrange the given circle into standard form, that is equation (ii) and get the required points.