Question

Find the radius and centre of the circle described by the equation
x2 + y2 − 2x − 4y +1=0
by writing it in the form (x − a)2 + (y − b)2 = c2 for suitable a, b and c.

Answer 1

${\underline{\sf{Given}}}$

Circle equation :x² + y² -2x -4y +1 = 0

${\underline{\sf{To\:Find}}}$

The radius of the circle

${\underline{\sf{Theory}}}$

Central form of equation of a circle

The equation of circle having cente (h,k) and radius r is

$\sf\:(x-h)+(y-k)=r{}^{2}$

${\underline{\sf{Solution}}}$

Given equation:

x² + y² -2x - 4y+1=0

Make this equation quadratic, by the completing square method .

$\sf(x {}^{2} -2 \times 1 \times x + 1 {}^{2} ) -{1}^{2} + (y {}^{2} - 2 \times y \times 2+ 2{}^{2} )- {2}^{2}+1=0$

$\implies \sf(x-1 ) {}^{2} - 1+ (y - 2) {}^{2}-4+1 = 0$

$\implies \sf(x -1) {}^{2} + (y - 2 ){}^{2} = 4$

$\implies \sf(x -1) {}^{2} + (y - 2) {}^{2} = 2 {}^{2}$

Now compare this equation with central form of a circle.

On comparing :

h=1

k= 2

and r = 2

Therefore,the radius of circle x² + y² + -2x-4y+1= 0 is 2 units .

$\rule{200}2$

More About circles :

•Genral equation of a circle :

The general equation of a circle is $\sf\:x{}^{2}+y{}^{2}+2gx+2fy+c=0$, where g,f,c are constants .

Answer 2

Aɴꜱᴡᴇʀ

☞ Radius of the circle = 2 units

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Gɪᴠᴇɴ

✭ The equation of the circle - x² + y² - 2x - 4y + 1 = 0

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Tᴏ ꜰɪɴᴅ

➤ The radius of the circle?

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Sᴛᴇᴘꜱ

❍ The Centre form of a circle with equation of the circle having center h,k and radius r is given by,

$\underline{\boxed{\red{\sf (x-h) + (y-k) = {r}^{2}}}}$

So first let's make the equation quadratic,

➳ (x²-2×1 × 1²) - 1² + (y²-2×y×2+2²) - 2²+1 = 0

➳ (x-1)² - 1 + (y-2)² - 4 +1 =0

➳ (x-1)² + (y-2)² = 4

➳ (x-1)² + (y-2)² = 2²

So we observe that here,

◕ h = 1

◕ r = 2

◕ k = 2

This the radius of the circle is 2 units

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