Question

secondary 3 | A Maths

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Megan Yifei

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Ok ive been staring at this qn for 2 days i still cant figure it out

Date Posted: 5 months ago
Views: 32
Eric Nicholas K
5 months ago
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.

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1st
Good evening Megan! Here are my workings for this question.
Date Posted: 5 months ago
J
5 months 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.