Add Maths - Circles
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It is given that P(0,3), Q(7,5), R(9,-2) and S are vertices of a square . A circle A passes through the points P, Q, R and S. Find the equation of this circle. Hence or otherwise find the equation of circle B which is the image of A under the reflection abot y-axis.
Need help with the 2nd part. Answer:x^2 + y^2 + x - y - 6 = 0
Answer for circle A(referencing purposes) x^2 +y^2 -9x-y-6=0
Thank you!
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confirmation that it is -9x - y - 6=0 ?
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Originally posted by anpanman:
It is given that P(0,3), Q(7,5), R(9,-2) and S are vertices of a square . A circle A passes through the points P, Q, R and S. Find the equation of this circle. Hence or otherwise find the equation of circle B which is the image of A under the reflection abot y-axis.
Need help with the 2nd part. Answer:x^2 + y^2 + x - y - 6 = 0
Answer for circle A(referencing purposes) x^2 +y^2 -9x-y-6=0
Thank you!
The answer for part 2 should be x^2+y^2+9x-y-6=0Because the center for circle A is (4.5, 0.5). Reflection of this point in the y-axis would be (-4.5,0.5) which is the center of circle B. The radius will remain the same.
Where the original circle is (x-a)^2+(y-b)^2= r^2
Center (a,b) and radius r.
The image of the reflection of (a,b) in the y-axis would be (-a,b) and the radius would still be r.
Giving you [x-(-a)]^2+(y-b)^2 = r ^2
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Originally posted by Mikethm:
The answer for part 2 should be x^2+y^2+9x-y-6=0Because the center for circle A is (4.5, 0.5). Reflection of this point in the y-axis would be (-4.5,0.5) which is the center of circle B. The radius will remain the same.
Where the original circle is (x-a)^2+(y-b)^2= r^2
Center (a,b) and radius r.
The image of the reflection of (a,b) in the y-axis would be (-a,b) and the radius would still be r.
Giving you [x-(-a)]^2+(y-b)^2 = r ^2
No. It should be -9x. I just checked through the first equation with my teacher. She said there's no problem with it. And this is also the answer given on the answer scheme. So yea, kinda reliable.
As what Mikethm has said, the centre is (4.5, 0.5)
You solve (x-4.5)^2 +(y-0.5)^2 = 53/2
x^2 - 9x + 81/4 + y^2 - y + 1/4 = 53/2
x^2 + y^2 - 9x - y - 6 =0
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By the way, Mikethm, I did use the approach you have specified. However the answer is x^2 +y^2 + 9x - y- 6 = 0. So is the answer sheet wrong? Anyway, thanks!
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Originally posted by anpanman:
No. It should be -9x. I just checked through the first equation with my teacher. She said there's no problem with it. And this is also the answer given on the answer scheme. So yea, kinda reliable.
As what Mikethm has said, the centre is (4.5, 0.5)
You solve (x-4.5)^2 +(y-0.5)^2 = 53/2
x^2 - 9x + 81/4 + y^2 - y + 1/4 = 53/2
x^2 + y^2 - 9x - y - 6 =0
Eh... I said that the 2nd equation should have +9x, not the 1st one.
I said that the 1st circle is (x-4.5)^2+(y-0.5)^2= r^2. Expanding that would give us -9x which is in agreement with your (teacher/answer sheet) solution for 1st circle.
The problem is that you said the 2nd circle is x^2 + y^2 + x - y - 6 = 0.
The correct coefficient of x is +9x for 2nd circle.
There is a program called geogebra. Here is what I did to create the attached image.1) I request the program to create a circle passing thru P, Q and R.
2) I request the program to reflect the circle in the y-axis.
The equations by the side are generated by the program.
