IIT JEE , AIEEE Forum - Mathematics , Algebra

nayan_dbb

Q.1)Let ?(x,y) be the equation of a circle . If ?(0, k ) =0 has equal roots k = 2 , 2 and ?( k ,0)
Has roots k= (4/5) , 5 then the centre of the circle is

a) ( 2 , (29/10) )
b) ( (29/10) , 2 )
c) ( -2 , (29/10) )
d) NONE OF THESE

Q.2)If a chord of the circle x^2 + y^2 = 8 makes equal intercepts of length ?a? on the coordinate axes then

a)| a | < 8
b) | a | < 4* 2 1/2
c) | a | < 4
d) | a | > 4

Q.3)The point Z in the complex plane describes a circle of radius 2 with centre at the origin ,then the point ( Z + (1/Z) ) describe

a)circle
b)parabola
c)ellipse
d)hyperbola

Q.4)The number of integral values of ?k? for which x^2 + y^2 + kx + (1- k)y + 5 = 0 is the equation of a circle whose radius cannot exceed 5 is

a)14
b)18
c)16
d)NONE OF THESE

Q.5)Two circles x^2 + y^2+ ax =0 and x^2 +y^2 = c^2 externally is

a) a + c = 0
b) a - c = 4
c) a^2 = c^2 + 1
d) NONE OF THESE

elessar_iitkgp

The first condition
f(0, k ) =0 has equal roots k = 2 , 2
means that it touches the Y axis at (0,2), ie, the Y axis is a tangent at (0,2).

The second condition
f( k ,0) = 0 has roots k= (4/5) , 5
means that the X axis cuts the circle at (4/5,0) and (5,0)

Let f(x,y) = x² + y² + 2gx + 2fy + c
Now you have three points which will give three equations in g,f and c.
Solve the system and find the center (-g,-f)

ramyadiamond

For q.2, ans is c.

The line is x+y=a, now solve it with the circle,

x² +y²=8

x²+(a-x)²=8

2x²-2ax+(a²-8)=0

Now since it is a chord, it cuts the cirle at two real points, hence the quadratic must have discriminant as positive.

4a²-8(a²-8)>0

so,

|a|<4

For q.4,

Centre is (-k/2, (k-1)/2)

now radius<= 5

radius² <=25

(k/2)²+(k-1/2)²-5<=25

On solving this condition, u'll get the required values.

For q.5, if the circles touch externally, then use the condition that the distance between the centrs must be equal to the sum of the radius of the circles given.

Hope this helps.

gopal_duggirala

for the fourth question

x^2+y^2+kx+(1-k)y+5=0

g^2+f^2-c=radius

here g=k

and f=(1-k)

c=5

so

k^2+(1-k)^2 - 5 is less or =5 (since radius cannot exceed 5)

solve this and get the answer

if u dont get nudge me, now am running out of time.....

please please please rate me...........

karthik2007

I didnt quite get the solution for q 4. Even i got that inequality, but how do u proceed after that?

ramyadiamond

u might get a quadratic after solving. Get its roots and observe the intervals they lie in. Now, consider the integral values lying in between these two extreme values, and the intergers nearest to them if they are coming out be fractions. Say, if u r getting interval as 1/2<k<4.6, then the integers to be considered are 1,2,3,4. Hope u got it.

karthik2007

o.O right... thanks.

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