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Sagot :
Given equation of the Circle is ,
[tex]\sf\implies x^2 + y^2 = 25 [/tex]
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 ,
[tex]\sf\implies (x-0)^2 +( y-0)^2 = 5 ^2[/tex]
Here we can say that ,
• Radius = 5 units
• Centre = (0,0)
Finding distance between the two points :-
[tex]\sf\implies Distance = \sqrt{ (0+4)^2+(2-0)^2} \\\\\sf\implies Distance = \sqrt{ 16 + 4 } \\\\\sf\implies Distance =\sqrt{20}\\\\\sf\implies\red{ Distance = 4.47 }[/tex]
Here we can see that the distance of point from centre is less than the radius.
Hence the point lies within the circle .
inside the circle
Step-by-step explanation:
we want to verify whether (-4,2) lies inside or outside or on the circle to do so recall that,
- if [tex]\displaystyle (x-h)^2+(y-k)^2>r^2[/tex] then the given point lies outside the circle
- if [tex]\displaystyle (x-h)^2+(y-k)^2<r^2[/tex] then the given point lies inside the circle
- if [tex]\displaystyle (x-h)^2+(y-k)^2=r^2[/tex] then the given point lies on the circle
step-1: define h,k and r
the equation of circle given by
[tex] \displaystyle {(x - h)}^{2} + (y - k) ^2= {r}^{2} [/tex]
therefore from the question we obtain:
- [tex] \displaystyle h= 0[/tex]
- [tex] \displaystyle k= 0[/tex]
- [tex] {r}^{2} = 25[/tex]
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
[tex] \displaystyle {( - 4 - 0)}^{2} + (2 - 0 {)}^{2} \stackrel {?}{ < } 25[/tex]
simplify parentheses:
[tex] \displaystyle {( - 4 )}^{2} + (2 {)}^{2} \stackrel {?}{ < } 25[/tex]
simplify square:
[tex] \displaystyle 16 + 4\stackrel {?}{ < } 25[/tex]
simplify addition:
[tex] \displaystyle 20\stackrel { \checkmark}{ < } 25[/tex]
hence,
the point (-4, 2) lies inside the circle
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