Eric Nicholas K's answer to Megan Yifei's Secondary 3 A Maths Singapore question.
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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.