The-radius-of-circle-having-minimum-area-which-touches-the-curve-y-4-x-2-and-the-lines-y-x-is-

Question Number 57522 by rahul 19 last updated on 07/Apr/19

T h e r a d i u s o f c i r c l e h a v i n g m i n i m u m

a r e a, w h i c h t o u c h e s t h e c u r v e y = 4 - x^{2}

a n d t h e l i n e s y =∣ x ∣ i s ?

Commented by rahul 19 last updated on 06/Apr/19

A n s : \frac{\sqrt{34} - 2}{2 \sqrt{2}} .

Commented by MJS last updated on 06/Apr/19

Commented by MJS last updated on 06/Apr/19

there are 5 possible circles

Answered by mr W last updated on 07/Apr/19

f o r m a x . c i r c l e :

\sqrt{2} R = 4 + R

\Rightarrow R = \frac{4}{\sqrt{2} - 1} = 4 (\sqrt{2} + 1)

Commented by rahul 19 last updated on 07/Apr/19

I^{'} v e e d i t t h e Q u e s .

m y a p o l o g i e s \dots

Commented by mr W last updated on 07/Apr/19

Answered by mr W last updated on 07/Apr/19

f o r m i n . c i r c l e :

e q n . o f c i r c l e x^{2} + {(y - \sqrt{2} r)}^{2} = r^{2}

s i n c e y = 4 - x^{2}

\Rightarrow 4 - y + y^{2} - 2 \sqrt{2} r y + 2 r^{2} = r^{2}

\Rightarrow y^{2} - (2 \sqrt{2} r + 1) y + r^{2} + 4 = 0

Δ = {(2 \sqrt{2} r + 1)}^{2} - 4 (r^{2} + 4) = 0

8 r^{2} + 4 \sqrt{2} r + 1 - 4 r^{2} - 16 = 0

4 r^{2} + 4 \sqrt{2} r - 15 = 0

\Rightarrow r = \frac{\sqrt{17} - \sqrt{2}}{2}

Commented by MJS last updated on 07/Apr/19

\frac{\sqrt{34} - 2}{2 \sqrt{2}} = \frac{\sqrt{2} \times \sqrt{2 \times 17} - 2 \sqrt{2}}{4} = \frac{2 \sqrt{17} - 2 \sqrt{2}}{4} =

= \frac{\sqrt{17} - \sqrt{2}}{2}

Commented by mr W last updated on 07/Apr/19

Commented by rahul 19 last updated on 07/Apr/19

T h a n k y o u s i r!

Answered by mr W last updated on 07/Apr/19

t h i r d p o s s i b i l i t y :

e q n . o f c i r c l e x^{2} + {(y + r)}^{2} = r^{2}

4 - y + y^{2} + 2 r y + r^{2} = r^{2}

y^{2} + (2 r - 1) y + 4 =

Δ = {(2 r - 1)}^{2} - 4 \times 4 = 0

2 r - 1 = 4

\Rightarrow r = \frac{5}{2}

Commented by mr W last updated on 07/Apr/19

Facebook Tweet Pin