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The equation of a circle is (x + 12)2 + (y + 16)2 = (r1)2, and the circle passes through the origin. the equation of the circle then changes to (x – 30)2 + (y – 16)2 = (r2)2, and the circle still passes through the origin. what are the values of r1 and r2?
a.r1 = 10 and r2 = 17
b.r1 = 10 and r2 = 34
c.r1 = 20 and r2 = 17
d.r1 = 20 and r2 = 34

Respuesta :

swafim
d.
let x=0 and y=0,
we get [tex] 12^2+16^2 =r _1^2 [/tex]
and [tex] 30^2+16^2 = r_2^2 [/tex]
so r1=20 and r2=34

Answer:

Option D is correct

Step-by-step explanation:

Given Equations of Circles:

Circle 1 - [tex](x+12)^2+(y+16)^2=(r_1)^2[/tex]

Circle 2 -  [tex](x-30)^2+(y-16)^2=(r_2)^2[/tex]

Both circles passes through origin.

To find: Values of [tex]r_1\:,\:r_2[/tex]

Coordinates of origin = ( 0 , 0 )

Circles passes through origin means x = 0 & y = 0 must satisfy the equation of circles.

So, Substituting x = 0 &  y = 0 in Eqn of Circle 1

we get

[tex](0+12)^2+(0+16)^2=(r_1)^2[/tex]

[tex](r_1)^2=12^2+16^2[/tex]

[tex](r_1)^2=144+256[/tex]

[tex](r_1)^2=400[/tex]

[tex]r_1=\sqrt{400}[/tex]

[tex]r_1=20[/tex]

Now, Substituting x = 0 &  y = 0 in Eqn of Circle 2

we get

[tex](0-30)^2+(0-16)^2=(r_2)^2[/tex]

[tex](r_2)^2=(-30)^2+(-16)^2[/tex]

[tex](r_2)^2=900+256[/tex]

[tex](r_2)^2=1156[/tex]

[tex]r_2=\sqrt{1156}[/tex]

[tex]r_2=34[/tex]

Therefore, Option D is correct .i.e., [tex]r_1=20\:\:,\:\:r_2=34[/tex]

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