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PIT_PIT [208]
2 years ago
12

Which is the equation of a circle whose center is at the origin and that passes through the point (3, 5)?

Mathematics
2 answers:
xxTIMURxx [149]2 years ago
7 0

Answer:

x² + y² = 34

Formula:

  • (x - h)² + (y - k)² = r²                    where (h, k) is the center

<u>Here find the radius using distance formula</u>:                → origin : (0, 0)

  • √(x2-x1)²+(y2-y1)²
  • √(3-0)²+(5-0)²
  • √9+25
  • √34

<u>Thus the equation of circle</u>:

  • (x - 0)² + (y - 0)² = (√34)²
  • (x - 0)² + (y - 0)² = 34
  • x² + y² = 34

AleksAgata [21]2 years ago
4 0

Answer:

H.  x^2+y^2=34

Step-by-step explanation:

<u>Equation of a circle (in standard form)</u>

<u />(x-h)^2+(y-k)^2=r^2

(where (h, k) is the center and r is the radius of the circle)

Given center = (0, 0):

\implies (x-0)^2+(y-0)^2=r^2

\implies x^2+y^2=r^2

As the circle passes through point (3, 5):

\implies 3^2+5^2=r^2

\implies r^2=34

<u>Final equation</u>

<u />x^2+y^2=34

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