IIT JEE , AIEEE Forum - Mathematics , Algebra

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.

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