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

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Good evening Megan! Here are my workings for this question.
J
J
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.