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Question: If the point \(\left( \lambda ,-\lambda \right) \) lies inside the circle \(x^{2}+y^{2}-4x+2y-8=0\),...

If the point (λ,−λ)\left( \lambda ,-\lambda \right) lies inside the circle x2+y2−4x+2y−8=0x^{2}+y^{2}-4x+2y-8=0, then find the range of λ\lambda.

Explanation

Solution

In this question it is given that if a point (λ,−λ)\left( \lambda ,-\lambda \right) lies inside the circle x2+y2−4x+2y−8=0x^{2}+y^{2}-4x+2y-8=0, then we have to find the range of λ\lambda. So to find the solution we have to know that if any point (a,b) lies inside the circle then we can write, a2+b2−4a+2b−8<0a^{2}+b^{2}-4a+2b-8<0 …………….(1).

Complete step-by-step answer:
Since it is given that point (λ,−λ)\left( \lambda ,-\lambda \right) lies inside the circle.
So by (1) we can write,
a2+b2−4a+2b−8<0a^{2}+b^{2}-4a+2b-8<0
⇒λ2+λ2−4λ+2(−λ)−8<0\Rightarrow \lambda^{2} +\lambda^{2} -4\lambda +2\left( -\lambda \right) -8<0
⇒2λ2−4λ−2λ−8<0\Rightarrow 2\lambda^{2} -4\lambda -2\lambda -8<0
⇒2λ2−6λ−8<0\Rightarrow 2\lambda^{2} -6\lambda -8<0
⇒2(λ2−3λ−4)<0\Rightarrow 2\left( \lambda^{2} -3\lambda -4\right) <0
⇒(λ2−3λ−4)<0\Rightarrow \left( \lambda^{2} -3\lambda -4\right) <0 [dividing both side by 2]
Now by middle term factorisation,
⇒λ2−4λ+λ−8<0\Rightarrow \lambda^{2} -4\lambda +\lambda -8<0
⇒λ(λ−4)+1(λ−4)<0\Rightarrow \lambda \left( \lambda -4\right) +1\left( \lambda -4\right) <0
Now by taking (λ−4)\left( \lambda -4\right) common,
⇒(λ−4)(λ+1)<0\Rightarrow \left( \lambda -4\right) \left( \lambda +1\right) <0........(2)
Whenever multiplication of two terms less than zero, i.e ab<0, then
Either a<0 and b>0,
Or, a>0 and b<0,

So by this we can write (2) as,
Either, (λ−4)<0\left( \lambda -4\right) <0 and (λ+1)>0\left( \lambda +1\right) >0
Which implies, λ<4\lambda <4 and λ>−1\lambda >-1.
So the range of λ\lambda is −1<λ<4-1<\lambda <4 .
Also this can be written as λ∈(−1,4)\lambda \in \left( -1,4\right)

Or, (λ−4)>0\left( \lambda -4\right) >0 and (λ+1)<0\left( \lambda +1\right) <0
Which implies, λ>4\lambda >4 and λ<−1\lambda <-1
So there is no common region by this above condition.
Thus the required solution is λ∈(−1,4)\lambda \in \left( -1,4\right) .

Note: To solve this type of question you should keep in mind that, choose proper conditions according to the position of a point. And also while solving any inequation always remember that whenever multiplication of two terms less than zero, i.e ab<0, then either, a<0 and b>0, Or, a>0 and b<0. And for sets or intervals the meaning of “and” is intersection.