Question

With the equation of the circle that has a center at the origin and passes through the point (1-6),​

Answer 1

Equation of a circle passing through origin

• x²+y²=r²

So

As (1,-6) lies on circle it will satisfy

• 1²+(-6)²=r^2
• 1+36=r²
• r²=37

Now

equation of the circle

• x²+y²=37

Answer 2

`\sf x^2+y^2=37`

Explanation:

Equation of a circle:
`\sf(x-h)^2+(y-k)^2=r^2`

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

Given the center is at (0, 0)

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

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

Given the circle passes through point (1, -6):

`\sf \implies (1)^2+(-6)^2=r^2`

`\sf \implies r^2=37`

Therefore, the equation of the circle is:

`\sf \implies x^2+y^2=37`