WB JEE 2024 | Circle Question 14 | Mathematics

WB JEE 2024

MCQ (Single Correct Answer)

-0.5

If two circles which pass through the points $$(0, a)$$ and $$(0,-a)$$ and touch the line $$\mathrm{y}=\mathrm{m} x+\mathrm{c}$$, cut orthogonally then

$$c^2=a^2\left(1+m^2\right)$$

$$c^2=a^2\left(2+m^2\right)$$

$$c^2=a^2\left(1+2 m^2\right)$$

$$2 c^2=a^2\left(1+m^2\right)$$

WB JEE 2022

MCQ (Single Correct Answer)

-0.25

A curve passes through the point (3, 2) for which the segment of the tangent line contained between the co-ordinate axes is bisected at the point of contact. The equation of the curve is

$$y = {x^2} - 7$$

$$x = {{{y^2}} \over 2} + 2$$

$$xy = 6$$

$${x^2} + {y^2} - 5x + 7y + 11 = 0$$

WB JEE 2022

MCQ (Single Correct Answer)

-0.25

If the equation of one tangent to the circle with centre at (2, $$-$$1) from the origin is 3x + y = 0, then the equation of the other tangent through the origin is

$$3x - y = 0$$

$$x + 3y = 0$$

$$x - 3y = 0$$

$$x + 2y = 0$$

WB JEE 2022

MCQ (Single Correct Answer)

-0.25

The side AB of $$\Delta$$ABC is fixed and is of length 2a unit. The vertex moves in the plane such that the vertical angle is always constant and is $$\alpha$$. Let x-axis be along AB and the origin be at A. Then the locus of the vertex is

$${x^2} + {y^2} + 2ax\sin \alpha + {a^2}\cos \alpha = 0$$

$${x^2} + {y^2} - 2ax - 2ay\cot \alpha = 0$$

$${x^2} + {y^2} - 2ax\cos \alpha - {a^2} = 0$$

$${x^2} + {y^2} - ax\sin \alpha - ay\cos \alpha = 0$$

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