Question

If the point (x, 4) lies on a circle whose center is at the origin and radius is 5 units then x =?
(a) $\pm 5$
(b) $\pm 3$
(c) $0$
(d) $\pm 4$

Answer 1

Hint: Write the standard equation of the circle given as ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ where (a, b) is the coordinates of the center of the circle and r is radius of the circle. Substitute the given values (a, b) = (0, 0) and r = 5 units to get the equation of the circle. Now, substitute the value of the given coordinates that is lying on the circle and satisfy the equation of this circle. Solve for the value of x to get the answer.

Complete step by step answer:
Here we have been provided with a circle of radius 5 units and whose center lies on the origin. We are asked to find the value of x if the point (x, 4) lies on this circle.
Now, we know that the equation of a circle in standard form is given as ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ where (a, b) is the coordinate of the center of the circle and r is radius of the circle. Since, the center lies on the origin and radius is 5 units so we have (a, b) = (0, 0) and r = 5. Substituting the values in the equation of the circle we get,
$\begin{align}
  & \Rightarrow {{\left( x \right)}^{2}}+{{\left( y \right)}^{2}}={{5}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=25 \\
\end{align}$

Now, it is given that the point (x, 4) lies on the above circle that means this point must satisfy the equation of the circle. So substituting the values of the given coordinates in the equation of the circle we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{4}^{2}}=25 \\
 & \Rightarrow {{x}^{2}}+16=25 \\
 & \Rightarrow {{x}^{2}}=9 \\
\end{align}$
Taking square root both the sides we get,
$\begin{align}
  & \Rightarrow x=\pm \sqrt{9} \\
 & \therefore x=\pm 3 \\
\end{align}$
So, the correct answer is “Option b”.

Note: Remember the equations of a circle in different forms like standard form, general form, polar form etc. An important form of a circle is given as ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ where $\left( -g,-f \right)$ is the center of the circle and $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ is the radius. Note that we have obtained two values of x. In case x = 3 we will have the point (x, 4) in the first quadrant and in case x = -3 we will have the point (x, 4) in the second quadrant.