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Circle question

2024 · 27 Jan · Shift 1 · Q42

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Mathematics
Circle
2024 · 27 Jan · Shift 1 · Q42

Circle question

2024 · 27 Jan · Shift 1 · Q42

JEE MainMathematicsCircleMCQ+4 / −1

Four distinct points

(2 k, 3 k),(1,0),(0,1)

and

(0,0)

lie on a circle for

k

equal to :

A
$\frac{3}{13}$
B
$\frac{2}{13}$
C
$\frac{5}{13}$
D
$\frac{1}{13}$

View written solutionFree

Correct answer: C

Let the general equation of the circle be $x^2+y^2+2gx+2fy+c=0.$

Since the circle passes through $(0,0)$ , $c=0.$ So the equation becomes $x^2+y^2+2gx+2fy=0.$

Since the circle passes through $(1,0)$ , $1+2g=0 \Rightarrow g=-\frac12.$
Since the circle passes through $(0,1)$ , $1+2f=0 \Rightarrow f=-\frac12.$

Hence the circle is $x^2+y^2-x-y=0.$

Now the point $(2k,3k)$ also lies on this circle. Substitute: $(2k)^2+(3k)^2-(2k)-(3k)=0$ $4k^2+9k^2-5k=0$ $13k^2-5k=0$ $k(13k-5)=0.$

So, $k=0 \quad \text{or} \quad k=\frac{5}{13}.$

But the four points are distinct. If $k=0$ , then $(2k,3k)=(0,0)$ , which is not distinct from the given point $(0,0)$ .

Therefore, $k=\frac{5}{13}.$

Checking options:

A: $\frac{3}{13}$ — incorrect
B: $\frac{2}{13}$ — incorrect
C: $\frac{5}{13}$ — correct
D: $\frac{1}{13}$ — incorrect

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