A circle passes through the points(-1,1),(0,6)and(5,5).Find the points on this circle the tangents at which are parallel to the straight line joining the origin to its centre. I solved the three equation using the three points to get the centre as (2,3). I recognize that the slope of the tangent would be 2/3, also that the origin lies on the circle, but I don't know how to solve it further, please help.
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1 Answer
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Hints:
Take $A (0,6) ; B (-1,1) ; C (5,5) $. Now find a line perpendicular to BC passing through its midpoint $(2,3) $. Find a line perpendicular to AC passing through its midpoint $(\frac52, \frac {11}2) $.
From these lines, you will get the centre and equation of the circle. Now, write the tangent in slope form with slope determined by the two points: origin and centre $(2,3) $.