The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the point
(a) (–5, 2)
(b) (2, –5)
(c) (5, –2)
(d) (–2, 5)
Answers
Answered by
0
Let,
r
>
0
is the radius of the circle in Question.
Because the circle touches the X-Axis at
(
3
,
0
)
, its
centre C must be,
C
=
C
(
3
,
±
r
)
.
Given that the point
P
=
P
(
1
,
−
2
)
lies on the circle, we have,
P
C
2
=
r
2
.
∴
(
1
−
3
)
2
+
(
±
r
+
2
)
2
=
r
2
...
[
∵
,
Distance Formula]
.
∴
4
+
(
r
2
±
4
r
+
4
)
−
r
2
=
0
.
∴
±
4
r
=
−
8
⇒
r
=
±
2
,
but, as
r
>
0
,
r
=
+
2
.
This means that, the centre is
C
=
C
(
3
,
±
2
)
,
and
,
r
=
2
.
Hence, the eqns. of the circles are,
(
x
−
3
)
2
+
(
y
±
2
)
2
=
2
2
,
∵
P
(
1
,
−
2
)
∈
(
x
−
3
)
2
+
(
y
−
2
)
2
=
2
2
∴
,
the circle under reference is
(
x
−
3
)
2
+
(
y
+
2
)
2
=
2
2
.
Of all the given points, only
(
5
,
−
2
)
satisfies the above eqn.
Hence,
(a)
(
5
,
−
2
)
is the Right
Similar questions