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secondary 3 | A Maths
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'Find the radius and centre'
Ok ive been staring at this qn for 2 days i still cant figure it out
Date Posted: 4 years ago
Views: 221
Will look at this question again by 4 am. Clue: perpendicular bisector of any chord or the line perpendicular to tangents will always pass centre.
y-coordinate of B is definitely 5.
We form one “normal to curve” equation at that tangent and then solve with y = 5 to find the x-coordinate of centre.
Radius is easy after thatz
y-coordinate of B is definitely 5.
We form one “normal to curve” equation at that tangent and then solve with y = 5 to find the x-coordinate of centre.
Radius is easy after thatz
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done
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Good evening Megan! Here are my workings for this question.
Date Posted:
4 years ago
Alternative way to find centre (should you forget about the properties of the perpendicular bisector of chord) :
Let coordinates of centre be (a,b)
Now radius is always the same from any point on the circle's circumference to the centre.
So length from each point to the centre is the same.
Use formula Length = √[(x2 - x1)² + (y2 - y1)²]
√[(0 - a)²+(5 + 2√21 - b)²] = √[(0 - a)²+(5 - 2√21 - b)²]
We can square both sides. Expression becomes
a² + (5 + 2√21 - b)² = a² + (5 - 2√21 - b)²
(5 + 2√21)² - 2b(5 + 2√21) + b² = (5 - 2√21)² - 2b(5 - 2√21) + b²
5² + 20√21 + 84 - 10b - 4b√21 = 5² - 20√21 + 84 - 10b + 4b√21
8b√21 = 40√21
b = 5
So the y coordinate is 5.
Let coordinates of centre be (a,b)
Now radius is always the same from any point on the circle's circumference to the centre.
So length from each point to the centre is the same.
Use formula Length = √[(x2 - x1)² + (y2 - y1)²]
√[(0 - a)²+(5 + 2√21 - b)²] = √[(0 - a)²+(5 - 2√21 - b)²]
We can square both sides. Expression becomes
a² + (5 + 2√21 - b)² = a² + (5 - 2√21 - b)²
(5 + 2√21)² - 2b(5 + 2√21) + b² = (5 - 2√21)² - 2b(5 - 2√21) + b²
5² + 20√21 + 84 - 10b - 4b√21 = 5² - 20√21 + 84 - 10b + 4b√21
8b√21 = 40√21
b = 5
So the y coordinate is 5.
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