Question

show that the hypotenuse of a right angle triangle can be a diameter of a circle passing through the vertex containing the perpendicular

Answer 1

If a circle touches the middle point of the hypotenuse of a right-angled triangle and passes through the middle point of the shorter side (b > a), what is its radius?

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Let C(0,0) B(a,0) and A(0,b) be the vertices of the triangle.

Let H(h,k) be the center of the circle. The circle passes through L(a2,0)and M(a2,b2) . It follows that H lies on the perpendicular bisector of LM. ∴ k=b4.AB has a slope −ba. Therefore HM must have slope ab.

therefore, k−b2h−a2=ab

This results in h=a2−b24a

r2=(h−a2)2+(k−0)2=b416a2+b216

r=b4aa2+b2−−−−−−√

Answer 2

Let the perpendicular sides be AB=7cm and BC=24cm

According to the Pythagoras theorem

AB2+BC2=AC2

72+242=49+576=625

AC=25

A circle passes through three vertices of a triangle is called circumcircle. Centre of this circle will be intersection of perpendicular bisectors of any of two sides of the triangle. Perpendicular bisector of side BCwill be parallel to side AB. In a triangle if we draw a parallel line to one side of the triangle through the mid point of another side that line will bisect the the third side. Perpendicular bisector to side BCwill bisect the side AC at O. Perpendicular bisector of AB also meet AC at O. Ois also intersecting point of perpendicular bisectors of ABand BC. So Ois the circle of the centre. It is the midpoint of AC. So AOis the radius of the circle and half of the ACthat is 12.5 cm.

Radius of the circle is 12.5cm