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Can anybody suggest a simpler method. Geometry construction problem.?
A circle with centre O is inscribed in square. Let P be a vertex with PA & PB tangent to the circle.
How to construct a circle with PA, PB tangents and touching the circle at Q.
I joined OP and divided PQ in the ratio
√2 : 1 = PO' : O'Q.
Then O' is the centre of the circle.
1 Answer
- LearnerLv 76 years agoFavorite Answer
Hello 'Shibashis',
i) At Q construct a line perpendicular to OP; this line is incidentally a tangent at
Q to the circle having center at O. Let this meet AP at C and BP at D.
ii) So required circle is a incircle of triangle CPD and we need to find the center of this circle.
This center, which is incenter, is the meeting point of the angle bisectors of the triangle CPD.
Already we have one angle bisector OP of the angle CPD.
So draw bisector for any one of the other two angles PCD or PDC.
Let it intersect PQ at O'
With O' as center and O'Q as radius draw the incircle.