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secondary 4 | A Maths
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Hi i need help with part iv, would also appreciate if you could help me check if i did part iii correctly as i'm unsure. thanks!
Date Posted: 4 years ago
Views: 254
Can I have whole qn? I think there are info I need before I can check iii. Part iv you just have to use the formula to rewrite the given eqn into the centre-radius form and check if the x-coordinate is the negative of that of C₁. If yes then the answer is yes, if it isn't it's not.
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Hi jiayang this isn't the answer lol but there's no way to reply to a comment by adding a photo so I hope you see this. thanks!
Date Posted:
4 years ago
V good, you have to use part ii to solve part iii. Hold on
Anw your part ii is wrong.
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Actually I didn't need your pic I was under the impression that you haven't learned calculus.
About part iv, I forgot about one more condition which is the radii must be equal also.
About part iv, I forgot about one more condition which is the radii must be equal also.
Date Posted:
4 years ago
thanks a lot :)
Another way to get the equation of the perpendicular bisector :
Equation of circle is (x - h)² + (y - k)² = r²
Substitute (4,2) and (10,8)
(4 - h)² + (2 - k)² = r²
(10 - h)² + (8 - k)² = r²
So (4 - h)² + (2 - k)² = (10 - h)² + (8 - k)² = r²
16 - 8h + h² + 4 - 4k + k² = 100 - 20h + h² + 64 - 16k + k²
12h + 12k = 144
h + k = 12
k = 12 - h
Let k = y and h = x,
y = 12 - x
This is the equation of the perpendicular bisector, which passes through the centre (h,k)
Equation of circle is (x - h)² + (y - k)² = r²
Substitute (4,2) and (10,8)
(4 - h)² + (2 - k)² = r²
(10 - h)² + (8 - k)² = r²
So (4 - h)² + (2 - k)² = (10 - h)² + (8 - k)² = r²
16 - 8h + h² + 4 - 4k + k² = 100 - 20h + h² + 64 - 16k + k²
12h + 12k = 144
h + k = 12
k = 12 - h
Let k = y and h = x,
y = 12 - x
This is the equation of the perpendicular bisector, which passes through the centre (h,k)
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