Question

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

Answer 1

• radius =
  `√(74)`
• center =
  `(4,-3)`

Explanation:

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

`\longrightarrow (x - 4)^2+(y+3)^2=74`

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

`\longrightarrow (x - h)^2+(y-k)^2=r^2`

where ,

• 
  `(x,y)` is a point on circle .
• 
  `(h,k)` is the centre of circle .
• 
  `r` is the radius of the circle .

We can rewrite the equation as ,

`\longrightarrow (x-4)^2+\{ y -(-3)\}^2=(√(74))^2\\`

Now on comparing to the standard form , we have ;

• radius =
  `r` =
  `√(74)`
• center =
  `(h,k)` =
  `(4,-3)`

Graph :-

`\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){$\sf √(74)$}\put(2,6){$\boxed{\sf \textcopyright \: RISH4BH }$}\end{picture}`

And we are done !

Answer 2

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 ;

`{:\implies \quad \sf (x-4)^(2)+(y+3)^(2)=74}`

Can be further written as ;

`{:\implies \quad \sf (x-4)^(2)+\{y-(-3)\}^(2)=({√(74)})^(2)}`

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

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