Circle

JEE Mathematics · 133 questions · Page 14 of 14 · Click an option or "Show Solution" to reveal answer

Q131

A circle passes through (−2, 4) and touches the y-axis at (0, 2). Which one of the following equations can represent a diameter of this circle?

A 4x + 5y − 6 = 0

B 2x − 3y + 10 = 0

C 3x + 4y − 3 = 0

D 5x + 2y + 4 = 0

Correct Answer

Option B

Solution

EF = perpendicular bisector of chord AB BG = perpendicular to y-axis Here C = center of the circle mid-point of chord AB, D = ( $-$ 1, 3) slope of AB =

{{4 - 2} \over { - 2 - 0}}

= $-$ 1 $\because$ EF $\bot$ AB $\therefore$ Slope of EF = 1 Equation of EF, y $-$ 3 = 1 (x + 1) $\Rightarrow$ y = x + 4 . . . .(i) Equation of BG y = 2 . . . . (ii) From equations (i) and (ii) x = $-$ 2, y = 2 since C be the point of intersection of EF and BG, therefore center, C = ( $-$ 2, 2) Now coordinates of center C satiesfy the equation 2x $-$ 3y + 10 = 0 Hence 2x $-$ 3y + 10 = 0 is the equation of the diameter

Q132

If a point P has co-ordinates (0,

-

2) and Q is any point on the circle, x2 + y2

-

-

y + 5 = 0, then the maximum value of (PQ)2 is :

{{25 + \sqrt 6 } \over 2}

B 14 +

5\sqrt 3

{{47 + 10\sqrt 6 } \over 2}

D 8 + 5

\sqrt 3

Correct Answer

Option B

Solution

Given that x2 + y2 $-$ 5x $-$ y + 5 = 0 $\Rightarrow$ (x $-$ 5/2)2 $-$

{{25} \over 4}

+ (y $-$ 1/2)2 $-$ 1/4 = 0 $\Rightarrow$ (x $-$ 5/2)2 + (y $-$ 1/2)2 = 3/2 on circle [ Q $\equiv$ (5/2 +

\sqrt {3/2}

cos Q,

{1 \over 2}

\sqrt {3/2}

sin Q)] $\Rightarrow$ PQ2 =

{\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2}

{\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}

$\Rightarrow$ PQ2 =

{{25} \over 2} + {3 \over 2} + 5\sqrt {3/2}

(cos Q + sinQ) = 14 + 5

\sqrt {3/2}

(cosQ + sinQ) $\therefore$ Maximum value of PQ2 = 14 + 5

\sqrt {3/2}

$\times$

\sqrt 2

= 14 + 5

\sqrt 3

Q133

If the circles x2 + y2

-

16x

-

20y + 164 = r2 and (x

-

4)2 + (y

-

7)2 = 36 intersect at two distinct points, then :

A r > 11

B 0 < r < 1

C r = 11

D 1 < r < 11

Correct Answer

Option D

Solution

Circles are x2 + y2 $-$ 16x $-$ 20y + 164 = r2 $\Rightarrow$ c1 (8, 10) and (x $-$ 4)2 + (y $-$ 7)2 = 36 they intersect at two distinct points

\left| {{r_1} - {r_2}} \right| < {c_1}{c_2} < {r_1} + {r_2}\left\{ {{c_1}{c_2} = \sqrt {16 + 9} = 5} \right\}

Now

\left| {r - 6} \right| < 5 < r + 6

\left| {r - 6} \right| < 5

$\Rightarrow$

- 5 < r - 6 < 5

$\Rightarrow$

1 < r < 11\,\,\,\,\,\,\,\,\,...(i)

5 < r + 6

- 1 < r\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)

from (i) and (ii) r

\in

(1, 11)

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