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Find the equation of circle that passes through the point $(2,2)$ and tangent to the line $x=1$ and $x=6$.

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closed as off-topic by Shailesh, Anurag A, Claude Leibovici, TastyRomeo, Watson Jan 8 '17 at 12:24

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Hint: Since the tangents are parallel, the centre of the circle lies at $O (\frac {7}{2}, k) $ and the radius equals $r= \frac {6-1}{2}=\frac {5}{2} $.

Can you take it from here?

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Since both your tangents $x=1$ and $x=6$ are parallel, the diameter of the circle is the distance between them i.e. is $5$ and the radius is $5/2$. Therefore the x coordinate of center is $1+5/2 = 7/2$

We know that equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$

Substituting known values gives $(x-7/2) + (y-b)^2 = (5/2)^2$

We also know that the circle passes through $(2, 2)$, we get,

$(2-7/2)^2 + (2-b)^2 = (5/2)^2$

$(2-b)^2 = (5/2)^2 - (-3/2)^2 = 4$

$(2-b) = \pm 2$

$b = 2 \pm 2$

$=>$ $b=0$ or $b=4$

So there are 2 possible circles

circle 1 circle 2

The 2 equations of the circles would be $(x-7/2)^2 + y = (5/2)^2$ and $(x-7/2)^2 + (y-4)^2 = (5/2)^2$

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  • $\begingroup$ Where did you get b=4? $\endgroup$ – Sheena Pascual Jan 8 '17 at 10:48
  • $\begingroup$ @SheenaPascual I updated my answer. $\endgroup$ – Shreyash S Sarnayak Jan 8 '17 at 12:50

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