Question

The area of a circle is given by the (πx² + 10πx + 25π) find the radius of the circle.​

Answer 1

Answer:

Area of circle

$\sf \pi x^2+10 \pi x +25\piπx </strong></p><p><strong>[tex]\sf \pi x^2+10 \pi x +25\piπx$

We have to find the radius of circle

Area of circle is given the quadratic form

now first find the value of x

By Quadratic formula

$\boxed{\sf \ x= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}} </p><p>$

Where ,

a= coefficient

$\sf{of \: x^2}$

b= coefficient of x

c= constant term

$\begin{gathered}\sf \dag\ \ \ \pi x^2+10\pi x+ 25 \pi \\ \\ \sf\bullet a= \pi \ \ \bullet b= 10\pi \ \ \bullet \ c= 25 \pi\end{gathered} </p><p>$

Put the values in the formula

$\begin{gathered}\mapsto\sf x= \dfrac{ -10\pi \pm \sqrt{(-10\pi )^2-4\times \pi \times 25\pi}}{2\times \pi}\\ \\ \\ \mapsto\sf x= \dfrac{-10\pi \pm \sqrt{100\pi^2 - 100\pi^2}}{2\pi}\\ \\ \\ \mapsto\sf x= \dfrac{-10 \pi \pm\sqrt{ 0}}{ 2\pi }\\ \\ \\ \mapsto\sf x= \dfrac{-10\pi \pm 0}{2 \pi}\\ \\ \\ \mapsto\sf x= \dfrac{-10\pi}{2\pi}+\dfrac{0}{2\pi} \ , \ \ \dfrac{-10\pi}{2\pi}-\dfrac{0}{2\pi}\\ \\ \\ \mapsto\sf x= \cancel{\dfrac{-10\pi}{2\pi}}\ \ \cancel{\dfrac{-10\pi}{2\pi}}\\ \\ \\ \mapsto\sf x= -5 ,-5 \\ \\ \mapsto\sf x= -5\end{gathered} </p><p></p><p>$

Now the Area of circle

$\begin{gathered}\sf \pi x^2+10\pi x+25\pi \\ \\ \mapsto\sf \pi( x^2+10x+25)\\ \\ \mapsto\sf \pi(x^2+5x+5x+25)\\ \\ \mapsto\sf \pi [x(x+5)+5(x+5)]\\ \\ \mapsto\sf \pi [(x+5)(x+5)]\\ \\ \sf\bullet\ \ Put \ value \ of \ x\\ \\ \mapsto\sf \pi[(-5+5)(-5+5)]\\ \\ \mapsto\sf \pi(0)\\ \\ \mapsto\sf Area \ of \ circle = 0\end{gathered}$

Now find the radius of circle

$</p><p>\boxed{\bigstar{\sf\ Area \ of \ circle = \pi r^2}} </p><p></p><p>$

$\begin{gathered}\mapsto\sf \pi r^2= 0\\ \\ \mapsto\sf r^2= \dfrac{0}{\pi}\\ \\ \mapsto\sf r= \sqrt{0}\\ \\ \mapsto\sf r= 0 \ unit\end{gathered} </p><p>$

$\underline{\bigstar{\sf\ Radius\ of \ circle= 0}} </p><p>$

✒Hence We are Done !!

Answer 2

The Radius of the circle = 0

Hope it helps