The radius of the larger circle lying in the first quadrant and t
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The radius of the larger circle lying in the first quadrant and touching the line 4x + 3y - 12 =0 and the co-ordinate axes,is

  • 5

  • 6

  • 7

  • 8


B.

6

Let the equation of the circle is

x2 + y2 + 2gx +2fy +c =0

Thus circle touch the coordinate axes and lying in the first quadrant, then 

g2 - c =0 and f2 - c = 0

 


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The parabola with directrix x + 2y - 1 = 0 and focus (1, 0) is

  • 4x2 - 4xy + y2 - 8x + 4y + 4 = 0

  • 4x2 + 4xy + y2 - 8x + 4y + 4 = 0

  • 4x2 + 5xy + y2 + 8x - 4y + 4 = 0

  • 4x - 4xy + y - 8x - 4y + 4 = 0


A.

4x2 - 4xy + y2 - 8x + 4y + 4 = 0

Let P(x, y) be any point on the parabola

By defination of Parabola PM = PS

x + 2y - 11+ 4 = x - 12 + y2On squaring both sides, we getx2 + 4y2 +1 + 4xy - 4y -2x                                = 5x2 + 1 - 2x + y24x2 + y2 - 8x + 4y - 4xy + 4 = 0


Let M be the foot of the perpendicular from a point P on the parabola y = 8(x - 3) onto its directrix and let S be the focus ofthe parabola. If  SPM is an equilateral triangle, then P is equal to

  • (43, 8)

  • (8, 43)

  • (9, 43)

  • (43, 9)


C.

(9, 43)

Given, that the SPM is equilateralAlso, given parabola is y2 = 8x - 3 focus of this parabola is S(5, 0) and vertex A(3, 0)

Let coordinate of P (h + at2, k + 2at) = P(3 + 2t2, 4t)

Then, coordinate of M(- 5, 4t)

We know that the side of this  equilateral trianle is 4a = 4 x 2 = 8

Now, PS = 8

3 + 2t2 - 52 + 4t2 = 8   2t2 - 22 + 4t2 = 8                       2t2 + 2 = 8                              2t2 = 6                                 t = 3

 P3 + 2 × 3, 4 × 3 = P9, 43


The length of the common chord of the circles of radii 15 and 20, whose centres are 25 unit of distance apart, is

  • 12

  • 16

  • 24

  • 25


C.

24

Given, r1 = 15 unitr2 = 20 un itC1C2 = 25unitsLet  AC2D = θThen, in right angled triangle ADC2,AD = r2sinθ ADr2 = sinθ     . . . iNow, in right angled triangle ADC1AD = r1sin90 - θ  ADr1 = cosθ     . . . iiOn squaring and adding eqs i and  ii, we getAD21r12 + 1r22 = 1 AD2r12 + r22r12r22 = 1 AD2 = r12r22r12 + r22 AD = 225 × 400225 × 400 AD = 15 × 2025 = 12Thus, length of common chord = 2AD= 2 × 12= 24 unit


Consider the circle x2 + y - 4x - 2y + c = 0 whose centre is A(2, 1). If the point P(10, 7) is such that the line segment PA meets the circle in Q with PQ = 5, then c is equal to

  • - 15

  • 20

  • 30

  • - 20


D.

- 20

Given equation of circle isx2 + y2 - 4x - 2y + c = 0whose centre is A2, 1

Now, AP = 2 - 102 + 1 - 72                 = - 82 + - 62 = 64 + 36 = 100                 = 10      AQ = AP - PQ = 10 - 5 = 5So, Q is the mid-point of AP = 10 + 22, 7 + 12 = 6, 4Since, Q lies on a circle 62 + 42 - 46 -24 + c = 0      36 + 16 - 24 - 8 + c = 0 20 + c = 0            c = - 20


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