JEE Main 2024 (Online) 1st February Morning Shift | Circle Question 13 | Mathematics

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 :

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

$5 x^2-y=-11$

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

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

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

Let a variable line passing through the centre of the circle $$x^2+y^2-16 x-4 y=0$$, meet the positive co-ordinate axes at the points $$A$$ and $$B$$. Then the minimum value of $$O A+O B$$, where $$O$$ is the origin, is equal to

JEE Main 2024 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-1

If one of the diameters of the circle $$x^2+y^2-10 x+4 y+13=0$$ is a chord of another circle $$\mathrm{C}$$, whose center is the point of intersection of the lines $$2 x+3 y=12$$ and $$3 x-2 y=5$$, then the radius of the circle $$\mathrm{C}$$ is :

3$$\sqrt2$$

$$\sqrt{20}$$

JEE Main 2024 (Online) 30th January Morning Shift

MCQ (Single Correct Answer)

-1

If the circles $$(x+1)^2+(y+2)^2=r^2$$ and $$x^2+y^2-4 x-4 y+4=0$$ intersect at exactly two distinct points, then

$$\frac{1}{2}<\mathrm{r}<7$$

$$3<\mathrm{r}<7$$

$$5<\mathrm{r}<9$$

$$0<\mathrm{r}<7$$

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