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1

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, and C1($$\alpha$$, $$\beta$$) and C2($$\gamma$$, $$\delta$$), C1 $$\ne$$ C2 are their centres, then |($$\alpha$$ + $$\beta$$) ($$\gamma$$ + $$\delta$$)| is equal to ___________.
Your Input ________

Answer

Correct Answer is 40

Explanation

Slope of line joining centres of circles = $${4 \over 3} = \tan \theta $$


$$ \Rightarrow \cos \theta = {3 \over 5},\sin \theta = {4 \over 5}$$

Now using parametric form

$${{x - 1} \over {\cos \theta }} = {{y - 2} \over {\sin \theta }} = \pm \,5$$

(x, y) = (1 + 5cos$$\theta$$, 2 + 5sin$$\theta$$)

($$\alpha$$, $$\beta$$) = (4, 6)

(x, y) = ($$\gamma$$, $$\delta$$) = (1 $$-$$ 5cos$$\theta$$, 2 $$-$$ 5sin$$\theta$$)

($$\gamma$$, s) = ($$-$$2, $$-$$2)

$$\Rightarrow$$ |($$\alpha$$ + $$\beta$$) ($$\gamma$$ + $$\delta$$)| = | 10x $$-$$ 4 | = 40
2

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
Let the equation x2 + y2 + px + (1 $$-$$ p)y + 5 = 0 represent circles of varying radius r $$\in$$ (0, 5]. Then the number of elements in the set S = {q : q = p2 and q is an integer} is __________.
Your Input ________

Answer

Correct Answer is 61

Explanation

$$r = \sqrt {{{{p^2}} \over 4} + {{{{(1 - p)}^2}} \over 4} - 5} = {{\sqrt {2{p^2} - 2p - 19} } \over 2}$$

Since, $$r \in (0,5]$$

So, $$0 < 2{p^2} - 2p - 19 \le 100$$

$$ \Rightarrow p \in \left[ {{{1 - \sqrt {239} } \over 2},{{1 - \sqrt {39} } \over 2}} \right) \cup \left( {{{1 + \sqrt {39} } \over 2},{{1 + \sqrt {239} } \over 2}} \right]$$

so, number of integral values of p2 is 61.
3

JEE Main 2021 (Online) 26th August Morning Shift

Numerical
The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d2 is equal to _____________.
Your Input ________

Answer

Correct Answer is 16

Explanation

Let point P(x, y)

A(0, 0), B(1, 0), C(0, 1), D(1, 1)

(PA)2 + (PB)2 + (PC)2 + (PD)2 = 18

$${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$$ = 18

$$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$$

$$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$$

$$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $$

$$ \Rightarrow {d^2} = 16$$
4

JEE Main 2021 (Online) 17th March Morning Shift

Numerical
The minimum distance between any two points P1 and P2 while considering point P1 on one circle and point P2 on the other circle for the given circles' equations

x2 + y2 $$-$$ 10x $$-$$ 10y + 41 = 0

x2 + y2 $$-$$ 24x $$-$$ 10y + 160 = 0 is ___________.
Your Input ________

Answer

Correct Answer is 1

Explanation

$${S_1}:{(x - 5)^2} + {(y - 5)^2} = 9$$

Centre (5, 5), r1 = 3

$${S_2}:{(x - 12)^2} + {(y - 5)^2} = 9$$

Centre (12, 5), r2 = 3



So (P1P2)min = 1

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