In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of r^2 + s^2?
(1) The circle has radius 2.
(2) The point (v2, -v2) lies on the circle.
Please explain. Thanks...
OA is D
coordinate - set10 Q23
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eqn of circle with center at origin is
x^2 +y^2 = r^2(radius square)
stmt1 : r^2 + s^2 =4 suff
stmt 2:
(v2, -v2) is this v^2, -v^2 or 2v,-2v, but in any case we dont know the radius & "v" here
so IMO ans should be A
if pt is (2,-2) then ans is D.
x^2 +y^2 = r^2(radius square)
stmt1 : r^2 + s^2 =4 suff
stmt 2:
(v2, -v2) is this v^2, -v^2 or 2v,-2v, but in any case we dont know the radius & "v" here
so IMO ans should be A
if pt is (2,-2) then ans is D.
Regards
Samir
Samir
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Hi,
Well the question is whether or not the data provided is sufficient to find out the value of r^2+s^2.
To know this you should know the equation of the circle.
equation of the circle is:
(x-h)^2+(y-k)^2= r^2
where:
(h,k) coordinates of the center of the circle in this case it is (0,0)
hence new equation is:
x^2+y^2= (radius)^2 ......eq 1
1) Using 1 radius is 2 and point on the circle is (r,s) so putting (r,s) in the new equation you get
r^2+s^2=4 hence 1 is sufficient to answer your question.
2) Using 2 we get
(v2)^2+(-v2)^2=(radius)^2
=> 2(V2)^2 = (radius)^2 ..eq2
since (r,s) also lie on the circle
r^2+s^2=(radius)^2 ..eq3
but from eq2 we know (radius)^2=2(V2)^2 substituting eq2 in eq3 we get
r^2+s^2=2(V2)^2
Here we see that there is still one variable v2 in the result.
So we can say answer is (a) that is 1 alone is sufficient to answer the question.
Hope this helps
Thanks
Komal
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Well the question is whether or not the data provided is sufficient to find out the value of r^2+s^2.
To know this you should know the equation of the circle.
equation of the circle is:
(x-h)^2+(y-k)^2= r^2
where:
(h,k) coordinates of the center of the circle in this case it is (0,0)
hence new equation is:
x^2+y^2= (radius)^2 ......eq 1
1) Using 1 radius is 2 and point on the circle is (r,s) so putting (r,s) in the new equation you get
r^2+s^2=4 hence 1 is sufficient to answer your question.
2) Using 2 we get
(v2)^2+(-v2)^2=(radius)^2
=> 2(V2)^2 = (radius)^2 ..eq2
since (r,s) also lie on the circle
r^2+s^2=(radius)^2 ..eq3
but from eq2 we know (radius)^2=2(V2)^2 substituting eq2 in eq3 we get
r^2+s^2=2(V2)^2
Here we see that there is still one variable v2 in the result.
So we can say answer is (a) that is 1 alone is sufficient to answer the question.
Hope this helps
Thanks
Komal
================
www.testinggeek.com
learn, share and keep learning