Question

If the centre of a circle is the origin and the radius is 6cm, find the          points where the axes intersect the circle.​

Answer 1

Answer:

The figure is shown in image

The figure is shown in image Now apply Pythagoras theorem in ΔOAB

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3 2

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3 2 OB=4c.m.

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3 2 OB=4c.m. similarly for ΔABD

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3 2 OB=4c.m. similarly for ΔABDBD=4c.m.

The figure is shown in image Now apply Pythagoras theorem in ΔOAB OA 2 =AB 2 +OB 2 5 2 =OB 2 +3 2 OB=4c.m. similarly for ΔABDBD=4c.m.Hence Distance b\w center of the circles is 8 c.m.