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$Q.(\lambda,\lambda + 1) \text {lies inside the curve } x= \sqrt{25-y^{2}} \space and \ y-axis, \ find \ \lambda$

Attempt- From the given equation, we deduce that the shape is circle.

For a point inside the circle, the equation of circle is- $x^2+y^2+2gx+2fy+c<0$

Plugging in the coordinates of the point, we get- $\lambda^{2}+(\lambda+1)^2+2 \lambda g+ 2( \lambda + 1) f+c<0$

Now, this is where I run into trouble, how do I find the values of g, f, and c.

I tried equating the value of x, with the radius formula to get their value but no luck.

$\sqrt {g^2+f^2-c} = \sqrt {25-y^2}$

Akshat Batra
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1 Answers1

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Hint:

The equation $x=\sqrt{5-y^2}$ represents the semicircle with positive $x$ and with center $(0,0)$ and radius $r=5$ .

The point $Q=(x,y)=(\lambda, 1+\lambda)$ is in this semicircle if: $$ \begin{cases} -5\le y\le 5\\ 0\le x \le \sqrt{5-y^2}\\ \end{cases} $$ that, for $x=\lambda$ and $y=1+\lambda$ becomes: $$ \begin{cases} -5\le (1+\lambda)\le 5\\ 0\le \lambda \le \sqrt{5-(1+\lambda)^2}\\ \end{cases} $$

can you solve this?

From: $$ \begin{cases} -5\le (1+\lambda)\\ (1+\lambda)\le 5\\ 0\le \lambda \\ \end{cases} $$ we have $0\le \lambda\le 3$ and, for $\lambda$in this interval, the other inequality becomes: $$ \lambda^2\le 25-(1+\lambda)^2 $$ with solution $-4\le \lambda \le 3$. So the solution of the starting system is $0\le \lambda \le 3$.

Emilio Novati
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