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Find the center and the radius of the circle.
(x-4)^2 + (y + 3)^2 = 74

Sagot :

VirαtKσhli

Answer:

  • radius = [tex] \sqrt{74} [/tex]
  • center = [tex](4,-3)[/tex]

Step-by-step explanation:

We would like to calculate the centre and the radius of the circle . The given equation is ,

[tex]\longrightarrow (x - 4)^2+(y+3)^2=74 [/tex]

As we know that the Standard equation of circle is given by ,

[tex]\longrightarrow (x - h)^2+(y-k)^2=r^2[/tex]

where ,

  • [tex](x,y)[/tex] is a point on circle .
  • [tex](h,k)[/tex] is the centre of circle .
  • [tex]r[/tex] is the radius of the circle .

We can rewrite the equation as ,

[tex]\longrightarrow (x-4)^2+\{ y -(-3)\}^2=(\sqrt{74})^2\\ [/tex]

Now on comparing to the standard form , we have ;

  • radius = [tex]r[/tex] = [tex]\sqrt{74}[/tex]
  • center = [tex](h,k)[/tex] =[tex](4,-3)[/tex]

Graph :-

[tex] \setlength{\unitlength}{7mm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(0,0){\circle*{0.2}}\put(0.2, - .1){(4,-3)}\put(-1.2,0){\vector(0,1){5}}\put(-1.2,0){\vector(0, - 1){5}}\put(-1,0.7){\vector(1,0){5}}\put(-1,0.7){\vector( - 1,0){5}}\put(0,0){\line(-1,0){2.3}}\put( - 1.2,-0.7){$\sf \sqrt{74}$}\put(2,6){$\boxed{\sf \textcopyright \: RISH4BH }$}\end{picture} [/tex]

And we are done !

View image VirαtKσhli

We are given the equation of circle (x - 4)² + (y + 3)² = 74 , but let's recall the standard equation of circle i.e (x - h)² + (y - k)² = r², where (h, k) is the centre of the circle and r being the radius ;

So, consider the equation of circle ;

[tex]{:\implies \quad \sf (x-4)^{2}+(y+3)^{2}=74}[/tex]

Can be further written as ;

[tex]{:\implies \quad \sf (x-4)^{2}+\{y-(-3)\}^{2}=({\sqrt{74}})^{2}}[/tex]

On comparing this equation with the standard equation of Circle, we will get, centre and radius as follows

  • Centre = (4, -3)
  • Radius = √74 units
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