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Question
Mathematics
If real numbers x and y satisfy x2+y2-16x+30y+280=0, then maximum value of (x2 + y2)1 / 2 is
Q. If real numbers
x
and
y
satisfy
x
2
+
y
2
−
16
x
+
30
y
+
280
=
0
,
then maximum value of
(
x
2
+
y
2
)
1/2
is
53
110
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A
15
B
20
C
25
D
30
Solution:
We have,
x
2
+
y
2
−
16
x
+
30
y
+
280
=
0
⇒
(
x
−
8
)
2
+
(
y
+
15
)
2
=
(
3
)
2
Centre
≡
(
8
,
−
15
)
, Radius
=
3
Now,
x
2
+
y
2
represents the distance of
(
x
,
y
)
from the origin.
Distance will be maximum along the diameter at the point
Q
.
max
(
x
2
+
y
2
)
=
OP
+
PQ
=
(
8
−
0
)
2
+
(
−
15
−
0
)
2
+
3
=
17
+
3
=
20