Math / Geometry

QuestionFind the equation of the circle with center at the origin that contains the point $(8,15)$ .

Studdy Solution

STEP 1

1. The circle is centered at the origin $(0,0)$ .
2. The circle passes through the point $(8,15)$ .

STEP 2

1. Recall the standard equation of a circle centered at the origin.
2. Use the given point to find the radius of the circle.
3. Substitute the radius into the standard equation.

STEP 3

Recall the standard equation of a circle centered at the origin:
$x^2 + y^2 = r^2$
where $r$ is the radius of the circle.

STEP 4

Use the given point $(8,15)$ to find the radius $r$ . The radius is the distance from the origin to the point $(8,15)$ , which can be calculated using the distance formula:
$r = \sqrt{(8-0)^2 + (15-0)^2}$ $r = \sqrt{8^2 + 15^2}$ $r = \sqrt{64 + 225}$ $r = \sqrt{289}$ $r = 17$

STEP 5

Substitute the radius $r = 17$ into the standard equation of the circle:
$x^2 + y^2 = 17^2$ $x^2 + y^2 = 289$
The equation of the circle is:
$\boxed{x^2 + y^2 = 289}$

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QuestionFind the equation of the circle with center at the origin that contains the point $(8,15)$ .

Studdy Solution

STEP 1

1. The circle is centered at the origin $(0,0)$ .
2. The circle passes through the point $(8,15)$ .

STEP 2

1. Recall the standard equation of a circle centered at the origin.
2. Use the given point to find the radius of the circle.
3. Substitute the radius into the standard equation.

STEP 3

Recall the standard equation of a circle centered at the origin:
$x^2 + y^2 = r^2$
where $r$ is the radius of the circle.

STEP 4

Use the given point $(8,15)$ to find the radius $r$ . The radius is the distance from the origin to the point $(8,15)$ , which can be calculated using the distance formula:
$r = \sqrt{(8-0)^2 + (15-0)^2}$ $r = \sqrt{8^2 + 15^2}$ $r = \sqrt{64 + 225}$ $r = \sqrt{289}$ $r = 17$

STEP 5

Substitute the radius $r = 17$ into the standard equation of the circle:
$x^2 + y^2 = 17^2$ $x^2 + y^2 = 289$
The equation of the circle is:
$\boxed{x^2 + y^2 = 289}$

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QuestionFind the equation of the circle with center at the origin that contains the point (8,15)(8,15)(8,15).

Studdy Solution

STEP 1

1. The circle is centered at the origin (0,0)(0,0)(0,0).2. The circle passes through the point (8,15)(8,15)(8,15).

STEP 2

1. Recall the standard equation of a circle centered at the origin.2. Use the given point to find the radius of the circle.3. Substitute the radius into the standard equation.

STEP 3

Recall the standard equation of a circle centered at the origin:x2+y2=r2 x^2 + y^2 = r^2 x2+y2=r2where r r r is the radius of the circle.

STEP 4

STEP 5

Substitute the radius r=17 r = 17 r=17 into the standard equation of the circle:x2+y2=172 x^2 + y^2 = 17^2 x2+y2=172 x2+y2=289 x^2 + y^2 = 289 x2+y2=289 The equation of the circle is:x2+y2=289 \boxed{x^2 + y^2 = 289} x2+y2=289​

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the equation of the circle with center at the origin that contains the point $(8,15)$ .

1. The circle is centered at the origin $(0,0)$ .
2. The circle passes through the point $(8,15)$ .

1. Recall the standard equation of a circle centered at the origin.
2. Use the given point to find the radius of the circle.
3. Substitute the radius into the standard equation.

Recall the standard equation of a circle centered at the origin:
$x^2 + y^2 = r^2$
where $r$ is the radius of the circle.

Substitute the radius $r = 17$ into the standard equation of the circle:
$x^2 + y^2 = 17^2$ $x^2 + y^2 = 289$
The equation of the circle is:
$\boxed{x^2 + y^2 = 289}$