Hi guys,
In the question "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 about y – axis."
Can someone help me with the part hence or otherwise? I've found that the equation of the circle is x^2 + y^2 -9x -y -6 =0
Thanks =)
Assuming your answer is correct, it can be written
(x - 9/2)^2 + (y - 1/2)^2 = 106/4
(check my numbers)
The reflection about the y axis would be
(x + 9/2)^2 + (y - 1/2)^2 = 106/4
Hi drwls,
Just wondering, does the word "under" in the question influence the question at all?
Thanks for your reply! =)
In this mathematics context, "Under the reflection about (an axis)" is the same as saying "after a reflection about (an axis)"
Mathematicions often us the preposition "under" to indicate carrying out an operation. It does not mean "below" in this case.
Thanks drwls for the good explanation =)
jack has 3/4 of a dollar. Maxwell has 1/10 of a dollar. Do they have more or less than $1.oo in all?
Sure! I can help you with finding the equation of circle B, the image of circle A under reflection about the y-axis.
To find the equation of circle B, we need to reflect the equation of circle A about the y-axis. Reflecting a point (x, y) about the y-axis is equivalent to changing the sign of the x-coordinate. So, for every point (x, y) on circle A, the corresponding point on circle B will have the coordinates (-x, y).
The equation of circle A is given as x^2 + y^2 - 9x - y - 6 = 0.
Now, let's substitute -x for x in the equation above to find the equation of circle B:
(-x)^2 + y^2 - 9(-x) - y - 6 = 0
x^2 + y^2 + 9x - y - 6 = 0
So, the equation of circle B is x^2 + y^2 + 9x - y - 6 = 0.
Therefore, the equation of circle B is x^2 + y^2 + 9x - y - 6 = 0.
I hope this helps! Let me know if you have any further questions.