Question

Find the equation of the circle which touches the x-axis at (a,0) and cuts off a chord of length p on the positive side of the y-axis. What is the equation of the circle when a=12 and p=10?

Answer 1

Hint:First, we have done some construction in the figure that we have joined the point in the circle O(12,y) with the chord AB given in the question with length 10 and marked it as C and also joined the point O with the chord point A. Then, by using the property of the circle that a perpendicular line from the center which is OC cuts the chord in two equal halves and by using the Pythagoras theorem for the right angles triangle OAC which states that the square of the hypotenuse h is equal to the sum of the square of perpendicular p and base b as${{h}^{2}}={{p}^{2}}+{{b}^{2}}$, we get the radius of the circle. Then, by using the equation of the circle where (h, k) is the centre and r is radius as${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$and substituting the values, we get the required equation of the circle.

Complete step by step answer:
In this question, we are supposed to find the equation of the circle which touches the x-axis at (a,0) and cuts off a chord of length p on the positive side of the y-axis when p is 10 and a is 12.
So, before proceeding for this, we must draw the circle which has the above conditions so that we can understand the question more appropriately as:

Now, we have done some construction in the figure that we have joined the point in the circle O(12,y) with the chord AB given in the question with length 10 and marked it as C and also joined the point O with the chord point A.
Now, we can see that x axis is acting like the tangent to the circle and the construction used above gives the point O as the centre of the circle.
Now, by using the property of the circle that a perpendicular line from the centre which is OC cuts the chord in two equal halves.
So, by using the above mentioned property, we get value of AC as:
$\begin{align}
  & AC=\dfrac{AB}{2} \\
 & \Rightarrow AC=\dfrac{10}{2} \\
 & \Rightarrow AC=5 \\
\end{align}$
Also, we already know the length of the segment OC as 12.
Now, by using the Pythagoras theorem for the right angles triangle OAC which states that the square of the hypotenuse h is equal to the sum of square of perpendicular p and base b as:
${{h}^{2}}={{p}^{2}}+{{b}^{2}}$
Now, by using the theorem, we get the values for triangle OAC by substituting h as AO, p as OC and b as AC, we get;
$\begin{align}
  & A{{O}^{2}}={{12}^{2}}+{{5}^{2}} \\
 & \Rightarrow A{{O}^{2}}=144+25 \\
 & \Rightarrow A{{O}^{2}}=169 \\
 & \Rightarrow AO=\sqrt{169} \\
 & \Rightarrow AO=13 \\
\end{align}$
Now, we can see clearly that AO is the radius of the circle which is 13 gives the value of the y coordinate of the circle which comes out to be 13.
Now, by using the equation of the circle where (h, k) is the centre and r is radius as:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$
Now, by substituting all the values of (h, k) as (12, 13) and radius as 13, we get:
$\begin{align}
  & {{\left( x-12 \right)}^{2}}+{{\left( y-13 \right)}^{2}}={{13}^{2}} \\
 & \Rightarrow {{\left( x-12 \right)}^{2}}+{{\left( y-13 \right)}^{2}}=169 \\
\end{align}$
Hence, we get the required equation of the circle as ${{\left( x-12 \right)}^{2}}+{{\left( y-13 \right)}^{2}}=169$.
Note:
Now, to solve these type of the questions we need to know some of the basics of the right angles triangle which states the Pythagoras theorem. Moreover, we should have the knowledge of the equation of the circle in standard form given by:
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$