Question

Does the point (-4, 2) lie inside or outside or on the circle x^2 + y^2 = 25?​
​

Answer 1

inside the circle

Explanation:

we want to verify whether (-4,2) lies inside or outside or on the circle to do so recall that,

1. if
   `\displaystyle (x-h)^2+(y-k)^2>r^2` then the given point lies outside the circle
2. if
   `\displaystyle (x-h)^2+(y-k)^2<r^2` then the given point lies inside the circle
3. if
   `\displaystyle (x-h)^2+(y-k)^2=r^2` then the given point lies on the circle

step-1: define h,k and r

the equation of circle given by

`\displaystyle {(x - h)}^(2) + (y - k) ^2= {r}^(2)`

therefore from the question we obtain:

• 
  `\displaystyle h= 0`
• 
  `\displaystyle k= 0`
• 
  `{r}^(2) = 25`

step-2: verify

In this case we can consider the second formula

the given points (-4,2) means that x is -4 and y is 2 and we have already figured out h,k and r² therefore just substitute the value of x,y,h,k and r² to the second formula

`\displaystyle {( - 4 - 0)}^(2) + (2 - 0 {)}^(2) \stackrel {?}{ < } 25`

simplify parentheses:

`\displaystyle {( - 4 )}^(2) + (2 {)}^(2) \stackrel {?}{ < } 25`

simplify square:

`\displaystyle 16 + 4\stackrel {?}{ < } 25`

simplify addition:

`\displaystyle 20\stackrel { \checkmark}{ < } 25`

hence,

the point (-4, 2) lies inside the circle

Answer 2

Given equation of the Circle is ,

`\sf\implies x^2 + y^2 = 25`

And we need to tell that whether the point (-4,2) lies inside or outside the circle. On converting the equation into Standard form and determinimg the centre of the circle as ,

`\sf\implies (x-0)^2 +( y-0)^2 = 5 ^2`

Here we can say that ,

• Radius = 5 units

• Centre = (0,0)

Finding distance between the two points :-

`\sf\implies Distance = √( (0+4)^2+(2-0)^2) \\\\\sf\implies Distance = √( 16 + 4 ) \\\\\sf\implies Distance =√(20)\\\\\sf\implies\red{ Distance = 4.47 }`

Here we can see that the distance of point from centre is less than the radius.

Hence the point lies within the circle .