Solveeit Logo

Question

Question: Four distinct points \(( 2 k , 3 k ) , ( 1,0 ) ( 0,1 )\) and \(( 0,0 )\) lie on a circle for....

Four distinct points (2k,3k),(1,0)(0,1)( 2 k , 3 k ) , ( 1,0 ) ( 0,1 ) and (0,0)( 0,0 ) lie on a circle for.

A

kI\forall k \in I

B

k<0k < 0

C

0<k<10 < k < 1

D

For two values of k

Answer

For two values of k

Explanation

Solution

General equation of circle is,

x2+y2+2gx+2fy+c=0x ^ { 2 } + y ^ { 2 } + 2 g x + 2 f y + c = 0

It passes through (0,0), (1, 0) and (0, 1); ∴ c=0c = 0

Now 2g+1=0g=122 g + 1 = 0 \Rightarrow g = - \frac { 1 } { 2 } and 2f+1=0f=122 f + 1 = 0 \Rightarrow f = \frac { - 1 } { 2 }

Hence equation of circle is x2+y2xy=0x ^ { 2 } + y ^ { 2 } - x - y = 0

Point (2k,3k)( 2 k , 3 k )lies on the circle

4k2+9k25k=04 k ^ { 2 } + 9 k ^ { 2 } - 5 k = 0

13k25k=013 k ^ { 2 } - 5 k = 0

k=0k = 0 or k=513k = \frac { 5 } { 13 }.