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

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

by fskilnik@GMATH » Tue Sep 11, 2018 5:39 am

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[GMATH practice question]

Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4

Answer: [spoiler]__(E)__[/spoiler]
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by fskilnik@GMATH » Wed Sep 12, 2018 5:29 am
fskilnik wrote:[GMATH practice question]

Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 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,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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