Solved on Jan 18, 2024

Determine if the point $(2,-3)$ lies on, inside, or outside the circle defined by the equation $x^{2}+y^{2}=9$ .

STEP 1

Assumptions
1. The equation of the circle is $x^{2}+y^{2}=9$ .
2. The point in question is $(2,-3)$ .
3. The center of the circle is at the origin $(0,0)$ .
4. The radius of the circle is the square root of the constant term on the right side of the equation, which is $\sqrt{9}=3$ .

STEP 2

To determine the location of the point $(2,-3)$ relative to the circle, we need to substitute the x and y coordinates of the point into the circle's equation and compare the result with the radius squared.
$x^{2}+y^{2} \, ? \, r^{2}$

STEP 3

Substitute the coordinates of the point $(2,-3)$ into the left side of the circle's equation.
$x^{2}+y^{2} = 2^{2}+(-3)^{2}$

STEP 4

Calculate the squares of the coordinates.
$x^{2}+y^{2} = 4+9$

STEP 5

Add the results to find the sum.
$x^{2}+y^{2} = 4+9 = 13$

STEP 6

Now compare the result with the square of the radius of the circle, which is $r^{2}=9$ .
$13 \, ? \, 9$

STEP 7

Since $13 > 9$ , the sum of the squares of the coordinates of the point $(2,-3)$ is greater than the radius squared of the circle.

STEP 8

Conclude that the point $(2,-3)$ lies outside the circle defined by the equation $x^{2}+y^{2}=9$ .
The point $(2,-3)$ lies outside the circle.

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Determine if the point $(2,-3)$ lies on, inside, or outside the circle defined by the equation $x^{2}+y^{2}=9$ .

STEP 1

Assumptions
1. The equation of the circle is $x^{2}+y^{2}=9$ .
2. The point in question is $(2,-3)$ .
3. The center of the circle is at the origin $(0,0)$ .
4. The radius of the circle is the square root of the constant term on the right side of the equation, which is $\sqrt{9}=3$ .

STEP 2

To determine the location of the point $(2,-3)$ relative to the circle, we need to substitute the x and y coordinates of the point into the circle's equation and compare the result with the radius squared.
$x^{2}+y^{2} \, ? \, r^{2}$

STEP 3

Substitute the coordinates of the point $(2,-3)$ into the left side of the circle's equation.
$x^{2}+y^{2} = 2^{2}+(-3)^{2}$

STEP 4

Calculate the squares of the coordinates.
$x^{2}+y^{2} = 4+9$

STEP 5

Add the results to find the sum.
$x^{2}+y^{2} = 4+9 = 13$

STEP 6

Now compare the result with the square of the radius of the circle, which is $r^{2}=9$ .
$13 \, ? \, 9$

STEP 7

Since $13 > 9$ , the sum of the squares of the coordinates of the point $(2,-3)$ is greater than the radius squared of the circle.

STEP 8

Conclude that the point $(2,-3)$ lies outside the circle defined by the equation $x^{2}+y^{2}=9$ .
The point $(2,-3)$ lies outside the circle.

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Determine if the point (2,−3)(2,-3)(2,−3) lies on, inside, or outside the circle defined by the equation x2+y2=9x^{2}+y^{2}=9x2+y2=9.

STEP 1

STEP 2

To determine the location of the point (2,−3)(2,-3)(2,−3) relative to the circle, we need to substitute the x and y coordinates of the point into the circle's equation and compare the result with the radius squared.x2+y2 ? r2x^{2}+y^{2} \, ? \, r^{2}x2+y2?r2

STEP 3

Substitute the coordinates of the point (2,−3)(2,-3)(2,−3) into the left side of the circle's equation.x2+y2=22+(−3)2x^{2}+y^{2} = 2^{2}+(-3)^{2}x2+y2=22+(−3)2

STEP 4

Calculate the squares of the coordinates.x2+y2=4+9x^{2}+y^{2} = 4+9x2+y2=4+9

STEP 5

Add the results to find the sum.x2+y2=4+9=13x^{2}+y^{2} = 4+9 = 13x2+y2=4+9=13

STEP 6

Now compare the result with the square of the radius of the circle, which is r2=9r^{2}=9r2=9.13 ? 913 \, ? \, 913?9

STEP 7

Since 13>913 > 913>9, the sum of the squares of the coordinates of the point (2,−3)(2,-3)(2,−3) is greater than the radius squared of the circle.

STEP 8

Conclude that the point (2,−3)(2,-3)(2,−3) lies outside the circle defined by the equation x2+y2=9x^{2}+y^{2}=9x2+y2=9.The point (2,−3)(2,-3)(2,−3) lies outside the circle.

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

Determine if the point $(2,-3)$ lies on, inside, or outside the circle defined by the equation $x^{2}+y^{2}=9$ .

To determine the location of the point $(2,-3)$ relative to the circle, we need to substitute the x and y coordinates of the point into the circle's equation and compare the result with the radius squared.
$x^{2}+y^{2} \, ? \, r^{2}$

Substitute the coordinates of the point $(2,-3)$ into the left side of the circle's equation.
$x^{2}+y^{2} = 2^{2}+(-3)^{2}$

Calculate the squares of the coordinates.
$x^{2}+y^{2} = 4+9$

Add the results to find the sum.
$x^{2}+y^{2} = 4+9 = 13$

Now compare the result with the square of the radius of the circle, which is $r^{2}=9$ .
$13 \, ? \, 9$

Since $13 > 9$ , the sum of the squares of the coordinates of the point $(2,-3)$ is greater than the radius squared of the circle.

Conclude that the point $(2,-3)$ lies outside the circle defined by the equation $x^{2}+y^{2}=9$ .
The point $(2,-3)$ lies outside the circle.