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secondary 3 | A Maths
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Date Posted: 3 years ago
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answer for Q14.
concept …
perpendicular bisector of any chord of a circle will always pass through the center of circle.
alternatively
perpendicular line from centre of circle, to any chord of the same circle, will bisect the chord into 2 equal parts
concept …
perpendicular bisector of any chord of a circle will always pass through the center of circle.
alternatively
perpendicular line from centre of circle, to any chord of the same circle, will bisect the chord into 2 equal parts
Date Posted:
3 years ago
for Q15, just apply the same method as for Q14
.... find the equation of perpendicular bisector of AB and solve the 2 equations as simultaneous equations to get center P.
the equation for the circle should be ...
(x-2)^2 + (y-1)^2 = 10
to check if a particular point lies within a circle, use Pythagoras theorem to find distance between the center and the given point.
if distance < radius, means point lies inside circle
if distance = radius, means point lies on circumference
if distance > radius, means point lies outside circle.
for point (-0.5,3), the distance is sqrt(41/4) which is just slightly more than the radius, so the point is outside.
.... find the equation of perpendicular bisector of AB and solve the 2 equations as simultaneous equations to get center P.
the equation for the circle should be ...
(x-2)^2 + (y-1)^2 = 10
to check if a particular point lies within a circle, use Pythagoras theorem to find distance between the center and the given point.
if distance < radius, means point lies inside circle
if distance = radius, means point lies on circumference
if distance > radius, means point lies outside circle.
for point (-0.5,3), the distance is sqrt(41/4) which is just slightly more than the radius, so the point is outside.
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