IIT-JEE 2010 - Career Point

[Q O(0, 0) and C(2, 2) lie on the same side of AB
Therefore, a
2 + b
2 – 1 < 0]
(2b + 2a – ab)
⇒ –
= 2
2 2
a + b
2
⇒ 2a + 2b – ab + 2 a + b = 0 ...(i)
Let P(h, k) be the circumcentre of ∆OAB. Since
∆ OAB is a right angled triangle. So its circumcentre
is the mid-point of AB.
a b
∴ h = and k = 2 2
⇒ a = 2h and b = 2k
From (i) and (ii), we get
4h + 4k – 4hk + 2
2
2
2
4 h + 4k = 0
⇒ h + k – hk + h + k = 0
So, the locus of P(h, k) is
2
2
...(ii)
x + y – xy + x + y = 0
But, the locus of the circumcentre is given to be
x + y – xy + k x + y = 0
Thus, the value of k is 1
2
13. Let C be any circle with centre (0, 2 ). Prove that at
most two rational points can be there on C.
(A rational points is a point both of whose
coordinates are rational numbers) [IIT-1997]
Sol. The equation of any circle C with centre (0, 2) is
given by
2
2
(x – 0) 2 + (y – 2) 2 = r 2 , where r is any positive real
number.
or, x 2 + y 2 – 2 2 y = r 2 – 2
If possible, let P(x 1 , y 1 ), Q(x 2 , y 2 ) and R(x 3 , y 3 ) be
three distinct rational points on circle C. Then,
2 2
x + y 2 2 y 1 = r 2 – 2 ...(ii)
x
1 1 −
2 2
2 2 −
2
+ y 2 2 y 2 = r 2 – 2 ...(iii)
2
x 3 + y3
− 2 2 y 3 = r 2 – 2 ...(iv)
We claim that at least two y 1 , y 2 , and y 3 are distinct.
For if y 1 = y 2 = y 3 , then P, Q and R lie on a line
parallel to x-axis and a line parallel to x-axis does not
cross the circle in more than two points. Thus, we
have either y 1 ≠ y 2 or, y 1 ≠ y 3 or, y 2 ≠ y 3 .
Subtracting (ii) from (iii) and (iv), we get
2 2 2 2
(x 2 + y 2)
– (x1 + y1
) – 2 2 (y 2 – y 1 ) = 0
and,
2 2 2 2
(x 3 + y3
) – (x1 + y1
) – 2 2 (y 3 – y 1 ) = 0
⇒ a 1 – 2b 1 = 0 and a 2 – 2b 2 = 0 ...(v)
2
where,
2 2 2 2
a 1 = (x 2 + y 2)
– (x1 + y1
) , b 1 = 2(y 2 – y 1 )
2 2 2 2
a 2 = (x 3 + y3
) – (x1 + y1
) , b 2 = 2(y 3 – y 1 )
Clearly, a 1 , a 2 , b 1 , b 2 are rational numbers as x 1 , x 2 ,
x 3 , y 1 , y 2 , y 3 are rational numbers.
Since either y 1 ≠ y 2 or, y 1 ≠ y 3
∴ Either b 1 ≠ 0 or, b 2 ≠ 0
If b 1 ≠ 0, then
a 1 – 2b 1 = 0 [From (v)]
a1
⇒ = 2,
b
1
a1
which is not possible because is a rational
b1
number and 2 is an irrational number.
If b 2 ≠ 0, then
a 2
a 2 – 2b 2 = 0 ⇒ = 2,
b
2
a 2
which is not possible because is a rational
b2
number and 2 is an irrational number.
Thus, in both the cases we arrive at a contradiction.
This means that our supposition is wrong. Hence,
there can be at most two rational points on circle C.
14. A rectangle PQRS has its side PQ parallel to the line
y = mx and vertices P, Q and S lie on the lines y = a,
x = b and x = –b, respectively. Find the locus of the
vertex R.
[IIT-1996]
Sol. Let the coordinates of R be (h, k). It is given that P
lies on y = a. So, let the coordinates of P be (x 1 , a).
Since PQ is parallel to the line y = mx. Therefore,
Slope of PQ = (Slope of y = mx) = m
1
And, Slope of PS = –
(Slope of y = mx)
1
= – [∴ PS ⊥ PQ]
m
Now, equation of PQ is
y – a = m(x – x 1 )
x = –b
x´ S
(0, – b)
(0, a)
R
O
y
y = 0
P
x = b
Q
(0, b)
x
...(i)
y´
It is given that Q lies on x = b. So, Q is the point of
intersection if (i) and x = b.
XtraEdge for IIT-JEE 16 DECEMBER 2009