2 equation of a circle question
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Hi, I need some help with the 2 qns below...
1. Given that a circle passes through the points P(3,5) and Q(-1,3) has radius 10^(1/2), find the equation of the circle.
The answer is x^2 + y^2 - 4x - 4y - 2 = 0.
2. Find the equation of the circle which passes through the points A(0,1), B(3,-2) and has its centre lying on the line y = x - 2
The answer is x^2 + y ^2 - 2x + 2y - 3 = 0.
Thanks a lot if you can help!
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(x-a)^2 + (y-b)^2 = c^2
x^2 - 2ax + a^2 + y^2 - 2by + b^2 = c^2
rearrange, x^2 + y^2 - 2ax + a^2 - 2by + b^2 = 10
subst in the x and y values from the 2 points given, solve simultaneous.
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Originally posted by Craappiboy:
(x-a)^2 + (y-b)^2 = c^2
x^2 - 2ax + a^2 + y^2 - 2by + b^2 = c^2
rearrange, x^2 + y^2 - 2ax + a^2 - 2by + b^2 = 10
subst in the x and y values from the 2 points given, solve simultaneous.
I have these steps somewhere in my answer. The problem is I have two answers in the end and I can't think of any reason to slap N.A. or something on the wrong one -
(x-a)^2 + (y-b)^2 = c^2
x^2 - 2ax + a^2 + y^2 - 2by + b^2 = c^2
rearrange, x^2 + y^2 - 2ax + a^2 - 2by + b^2 = 10
sub in P(3,5)
9 + 25 - 6a + a^2 - 10b + b^2 = 10
a^2 - 6a + b^2 - 10b = -24 - (1)
sub in Q(-1,3)
1 + 2a + a^2 + 9 - 6b + b^2 = 10
a^2 + 2a + b^2 - 6b = 0 - (2)
(1) - (2)
(-6a-10b) - (2a-6b) = -24
- 8a - 4b = -24
8a + 4b = 24
b = (24 - 8a) / 8 = 3 - a
subst into (2): a^2 + 2a + b^2 - 6b = 0
a^2 + 2a + 9 - 6a + a^2 - 18 + 6a = 0
2a^2 + 2a - 9 = 0
dam dunno how to continue
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err.. i think u cant cont because you sub the values u got into the same equation u derived them from.. u supposed to sub into (x-h)^2+(y-k)^2=r^2, not (x-h)^2+(y-k)^2=(x-h)^2+(y-k)^2
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Originally posted by undestu:
Hi, I need some help with the 2 qns below...
1. Given that a circle passes through the points P(3,5) and Q(-1,3) has radius 10^(1/2), find the equation of the circle.
The answer is x^2 + y^2 - 4x - 4y - 2 = 0.
2. Find the equation of the circle which passes through the points A(0,1), B(3,-2) and has its centre lying on the line y = x - 2
The answer is x^2 + y ^2 - 2x + 2y - 3 = 0.
Thanks a lot if you can help!
Lazy to do but give you the steps ok?
1.
Step 1. Find equation of perpendicular bisector of PQ
Step 2. Sub P (can be Q also...) into general equation of circle.
Step 3. Since the equation of bisector of PQ passes thru the center(C). Sub (a,b) into this.
Step 4. Solve the equations from step 2 & 3 simultaneously.
Step 5. Test if CQ (since you got it from using P) is correct for the two values you got in step 4. Reject the wrong one and write down equation of circle.
2.
Step 1. Find equation of perpendicular bisector of AB
Step 2. Since the equation of the perpendicular bisector and y=x-2 intersect at the center, solve simultaneous and get the coordinates of the center(C).
Step 3 Find the lenght of radius using either AC or BC.
Step 4 Write down equation of circle.
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i dont understand leh. i tot o lvl emath is over?
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Originally posted by davidche:
i dont understand leh. i tot o lvl emath is over?
This topic is not E Maths. It is A Maths.
TS probably a 2008 Sec 3 student.
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Originally posted by Mikethm:
Lazy to do but give you the steps ok?
1.
Step 1. Find equation of perpendicular bisector of PQ
Step 2. Sub P (can be Q also...) into general equation of circle.
Step 3. Since the equation of bisector of PQ passes thru the center(C). Sub (a,b) into this.
Step 4. Solve the equations from step 2 & 3 simultaneously.
Step 5. Test if CQ (since you got it from using P) is correct for the two values you got in step 4. Reject the wrong one and write down equation of circle.
2.
Step 1. Find equation of perpendicular bisector of AB
Step 2. Since the equation of the perpendicular bisector and y=x-2 intersect at the center, solve simultaneous and get the coordinates of the center(C).
Step 3 Find the lenght of radius using either AC or BC.
Step 4 Write down equation of circle.
Well thanks for trying to helping but it seems like something is wrong with question 2 and the answerlist. The perpendicular bisector of AB in Q2 is y=x-2 and both equations in Q1 are correct (according to a few maths teachers in school) so im at another topic now... can't help but feel something isn't right though
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Originally posted by undestu:
Well thanks for trying to helping but it seems like something is wrong with question 2 and the answerlist. The perpendicular bisector of AB in Q2 is y=x-2 and both equations in Q1 are correct (according to a few maths teachers in school) so im at another topic now... can't help but feel something isn't right though
Q1 simply have 2 possible solutions. (0,6) or (2,2) as the center.
Q2 is flawed. From ShingLee Textbook right? My bad for just going thru logics and assuming the question is correct.
Despite the occasional flaws, the ShingLee Text is much better than the TradePac one. :)