Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin

On the vertical axis, x=0.
So, the intersection points of the circle with vertical axis are (0,\(y_{1}\)) and (0,\(y_{2}\))

Now, put x=0 in the equation of the given circle to get the values of \(y_{1}\) and ,\(y_{2\)

Hence, \((0-3)^2+(y-1)^2=13\)
or, \(9+(y-1)^2=13\)
or, \((y-1)^2=4\)
Or, y-1=2, y-1=-2
Or, \(y_{1}\)=3, \(y_{2}\)=-1

p=3-(-1)=4

Ans.(E)

OA: E

The equation of circle : \((x-3)^2 + (y-1)^2 = 13\)

Centre of circle :\((3,1)\)

\({Radius}^2 \quad=13\)


Using Pythagoras theorem, we get

Required Chord length \(= 2* \sqrt{{{Radius}^2-3^2}}= 2* \sqrt{{13-9}}=2* \sqrt{{4}}=4\)

In cases like these, to find distance, how will we choose coordinates as Y1 and Y2? Like in this problem, while taking the difference b/w y-coordinates, how have you chosen the y coordinates? Cant it be -1-3=-4?

\(? = p\)

\(\left\{ \begin{gathered}\\
{\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} = 13 \hfill \\\\
x = 0 \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\left( {y - 1} \right)^2} = 4\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}\\
y - 1 = - 2 \hfill \\\\
\,\,\,\,\,{\text{OR}} \hfill \\\\
y - 1 = 2 \hfill \\ \\
\end{gathered} \right.\)

\(\left\{ \begin{gathered}\\
x = 0\,\,\,;\,\,\,y - 1 = - 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{point}}\,\,\,A = \,\,\left( {0, - 1} \right) \hfill \\\\
\,\,\,\,\,{\text{OR}} \hfill \\\\
x = 0\,\,\,;\,\,\,y - 1 = 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{point}}\,\,B = \left( {0,3} \right) \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\)

\(? = p = {\text{dist}}\left( {A,B} \right) = 4\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.