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  • If is a point on a diameter of the circle <img alt="x^{2}+y^{2}=4" src="https://entrancecorner.oncodecogs.c

If (a, 0) is a point on a diameter of the circle x^{2}+y^{2}=4, then x^{2}-4 x-a^{2}=0 has

Option: 1

Exactly one real root in (-5,-1)


Option: 2

Exactly one real root in (2,5)


Option: 3

Distinct roots less than -1


Option: 4

 Distinct roots greater than 5


Answers (1)

best_answer

Since (a, 0) is a point on the diameter of the circle x^{2}+y^{2}=4,

So, maximum value of \mathrm{a}^{2} is 4

\text{Let} \: f(x)=x^{2}-4 x-a^{2}

\text{Clearly} \: \mathrm{f}(-1)=5-\mathrm{a}$ is $1>0

f(b)=-\left(a^{2}+4\right)<0
f(0)=-a^{2}<0\: and \: f(5)=5-a^{2}>0
So graph of \mathrm{f}(\mathrm{x}) will be as shown

Hence (b) is the correct answer.

Posted by

Devendra Khairwa

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