Question

There is one and only one circle passing through three given non collinear points . Prove this theorem​

Answer 1

Step-by-step explanation:

THEOREM 10.5 ( ONLY ONE CIRCLE PASSING THROUGH THREE NON COLLINEAR POINTS) CIRCLES |

Answer 2

${\huge{\sf{\green{\underline{\underline{Solution:-}}}}}}$

${\blue{\sf\underline{Question:-}}}$

★ There is one and only one circle passing through three given non collinear points . Prove this theorem

${\blue{\sf\underline{Given:-}}}$

★ Three noncollinear point A , B , C

${\blue{\sf\underline{To\:Prove:-}}}$

★ There is one and only one circle passing through A , B , C

${\blue{\sf\underline{Construction:-}}}$

★ Join AB and BC. Draw the perpendicular bisector PQ and RS of AB and BC respectively. Let PQ and RS intersect at O.

Join OA , OB and OC.

${\blue{\sf\underline{Proof:-}}}$

Since O light on the perpendicular bisector of AB, we have

OA = OB ...(i)

Again, O lies on the perpendicular bisector of BC

∴ OB = OC. ...(ii)

Thus, OA = OB = OC = r (say) [from the (i) and (ii)]

With O as centre and radius r draw a circle C(O,r).

Clearly, C(O,r) passes through A, B and C

We shall show that this is the only circle passing through A, B, C.

If possible, let there be another circle C(O' , s), passing through the points A, B, C.

Then, O' will lie on the perpendicular bisector PQ and RS of AB and BC respectively. Also PQ and RS intersects at O.

Since two line cannot intersect at more than one point, so O' must coincide with O.

Since OA = r , O'A = s and O coincides with O', we must have r = s.

∴ C(O,r) ≅ C(O',s).

Hence, there is one and only one circle, passing through three noncollinear points A, B, C.

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