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JEE Advanced 2021 Paper 2 Online

Numerical
Consider the region R = {(x, y) $$\in$$ R $$\times$$ R : x $$\ge$$ 0 and y2 $$\le$$ 4 $$-$$ x}. Let F be the family of all circles that are contained in R and have centers on the x-axis. Let C be the circle that has largest radius among the circles in F. Let ($$\alpha$$, $$\beta$$) be a point where the circle C meets the curve y2 = 4 $$-$$ x.

The value of $$\alpha$$ is ___________.
Your Input ________

Answer

Correct Answer is 2.00

Explanation

Given, x $$\ge$$ 0, y2 $$\le$$ 4 $$-$$ x

Let equation of circle be

(x $$-$$ h)2 + y2 = h2 .... (i)


Solving Eq. (i) with y2 = 4 $$-$$ x, we get

x2 $$-$$ 2hx + 4 $$-$$ x = 0

$$\Rightarrow$$ x2 $$-$$ x(2h + 1) + 4 = 0 .... (ii)

For touching/tangency, Discriminant (D) = 0

i.e. (2h + 1)2 = 16 $$\Rightarrow$$ 2h + 1 = $$\pm$$ 4

$$\Rightarrow$$ 2h = $$\pm$$ 4 $$-$$ 1

$$\Rightarrow$$ $$h = {3 \over 2},h = {{ - 5} \over 2}$$ (Rejected) because part of circle lies outside R. So, $$h = {3 \over 2}$$ = radius of circle (c).

Putting h = 3/2 in Eq. (ii),

x2 $$-$$ 4x + 4 = 0 $$\Rightarrow$$ (x $$-$$ 2)2 = 0 $$\Rightarrow$$ x = 2

So, $$\alpha$$ = 2
2

JEE Advanced 2021 Paper 2 Online

Numerical
Consider the region R = {(x, y) $$\in$$ R $$\times$$ R : x $$\ge$$ 0 and y2 $$\le$$ 4 $$-$$ x}. Let F be the family of all circles that are contained in R and have centers on the x-axis. Let C be the circle that has largest radius among the circles in F. Let ($$\alpha$$, $$\beta$$) be a point where the circle C meets the curve y2 = 4 $$-$$ x.

The radius of the circle C is ___________.
Your Input ________

Answer

Correct Answer is 1.50

Explanation

Given, x $$\ge$$ 0, y2 $$\le$$ 4 $$-$$ x

Let equation of circle be

(x $$-$$ h)2 + y2 = h2 .... (i)


Solving Eq. (i) with y2 = 4 $$-$$ x, we get

x2 $$-$$ 2hx + 4 $$-$$ x = 0

$$\Rightarrow$$ x2 $$-$$ x(2h + 1) + 4 = 0 .... (ii)

For touching/tangency, Discriminant (D) = 0

i.e. (2h + 1)2 = 16 $$\Rightarrow$$ 2h + 1 = $$\pm$$ 4

$$\Rightarrow$$ 2h = $$\pm$$ 4 $$-$$ 1

$$\Rightarrow$$ $$h = {3 \over 2},h = {{ - 5} \over 2}$$ (Rejected) because part of circle lies outside R. So, $$h = {3 \over 2}$$ = radius of circle (C).
3

JEE Advanced 2020 Paper 2 Offline

Numerical
Let O be the centre of the circle x2 + y2 = r2, where $$r > {{\sqrt 5 } \over 2}$$. Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is 2x + 4y = 5. If the centre of the circumcircle of the triangle OPQ lies on the line x + 2y = 4, then the value of r is .............
Your Input ________

Answer

Correct Answer is 2

Explanation

As we know that the equation of family of circles passes through the points of intersection of given circle x2 + y2 = r2 and line PQ : 2x + 4y = 5 is,

(x2 + y2 $$-$$ r2) + $$\lambda $$(2x + 4y $$-$$ 5) = 0 ......(i)

Since, the circle (i) passes through the centre of circle

x2 + y2 = r2,

So, $$-$$ r2 $$-$$ 5$$\lambda $$ = 0

or 5$$\lambda $$ + r2 = 0 ....(ii)

and the centre of circle (i) lies on the line x + 2y = 4, so centre ($$-$$ $$\lambda $$, $$-$$ 2$$\lambda $$) satisfy the line x + 2y = 4.

Therefore, $$-$$$$\lambda $$ $$-$$4$$\lambda $$ = 4

$$ \Rightarrow $$ $$-$$5$$\lambda $$ = 4

$$ \Rightarrow $$ r2 = 4 {from Eq. (ii)}

$$ \Rightarrow $$ r = 2
4

JEE Advanced 2019 Paper 1 Offline

Numerical
Let the point B be the reflection of the point A(2, 3) with respect to the line $$8x - 6y - 23 = 0$$. Let $$\Gamma_{A} $$ and $$\Gamma_{B} $$ be circles of radii 2 and 1 with centres A and B respectively. Let T be a common tangent to the circles $$\Gamma_{A} $$ and $$\Gamma_{B} $$ such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is .................
Your Input ________

Answer

Correct Answer is 10

Explanation

According to given information the figure is as following



From the figure,

$$AC = {2 \over {\sin \theta }}$$ ....(i)

$$ \because $$ $$\sin \theta = {1 \over {CB}}$$ (from $$\Delta $$CPB) .... (ii)

and $$\sin \theta = {2 \over {AC}} = {2 \over {CB + AB}}$$ (from $$\Delta $$CQA) ....(iii)

$$ \because $$ AB = AM + MB = 2AM [$$ \because $$ AM = MB]

$$ = 2{{\left| {(8 \times 2) - (6 \times 3) - 23} \right|} \over {\sqrt {64 + 36} }}$$

$$ = {{2 \times 25} \over {10}} = 5.00$$

From Eqs. (ii) and (iii), we get

$$\sin \theta = {1 \over {CB}} = {2 \over {CB + AB}}$$

$$ \Rightarrow {1 \over {CB}} = {2 \over {CB + 5}}$$ [$$ \because $$ AB = 5]

$$ \Rightarrow CB + 5 = 2CB \Rightarrow CB = 5 = {1 \over {\sin \theta }}$$

From the Eq. (i), we get

$$AC = {2 \over {\sin \theta }} = 2 \times 5 = 10.00$$

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