JEE Main 2024 (Online) 4th April Evening Shift | Circle Question 10 | Mathematics

Let $$\mathrm{C}$$ be a circle with radius $$\sqrt{10}$$ units and centre at the origin. Let the line $$x+y=2$$ intersects the circle $$\mathrm{C}$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. Let $$\mathrm{MN}$$ be a chord of $$\mathrm{C}$$ of length 2 unit and slope $$-1$$. Then, a distance (in units) between the chord PQ and the chord $$\mathrm{MN}$$ is

$$3-\sqrt{2}$$

$$2-\sqrt{3}$$

$$\sqrt{2}-1$$

$$\sqrt{2}+1$$

JEE Main 2024 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

A square is inscribed in the circle $$x^2+y^2-10 x-6 y+30=0$$. One side of this square is parallel to $$y=x+3$$. If $$\left(x_i, y_i\right)$$ are the vertices of the square, then $$\Sigma\left(x_i^2+y_i^2\right)$$ is equal to:

152

148

156

160

JEE Main 2024 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

-1

Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $\mathrm{P}$ and $\mathrm{Q}$. Then, the length of $\mathrm{PQ}$ is :

$\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}$ intersect at two distinct points, is $\mathrm{R}-[\mathrm{a}, \mathrm{b}]$, then the point $(8 \mathrm{a}+12,16 \mathrm{~b}-20)$ lies on the curve :

$x^2+2 y^2-5 x+6 y=3$

$5 x^2-y=-11$

$x^2-4 y^2=7$

$6 x^2+y^2=42$

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