Question

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

Answer 1

H.
`x^2+y^2=34`

Explanation:

Equation of a circle (in standard form)

`(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`

Final equation

`x^2+y^2=34`

Answer 2

x² + y² = 34

Formula:

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

Here find the radius using distance formula: → origin : (0, 0)

• √(x2-x1)²+(y2-y1)²

• √(3-0)²+(5-0)²

• √9+25

• √34

Thus the equation of circle:

• (x - 0)² + (y - 0)² = (√34)²

• (x - 0)² + (y - 0)² = 34

• x² + y² = 34