Question Find the equation of a circle with center at $(-3, 6)$ given that the line $3x - y - 5 = 0$ is tangent to the circle.

Studdy Solution

STEP 1

Assumptions
1. The equation of the line tangent to the circle is $3x - y - 5 = 0$ .
2. The center of the circle is $(-3,6)$ .
3. The radius of the circle is the perpendicular distance from the center of the circle to the tangent line.
4. The general equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$ .

STEP 2

First, we need to find the perpendicular distance from the center of the circle to the tangent line. This distance will be the radius of the circle.

STEP 3

The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula:
$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

STEP 4

Plug in the values for the center of the circle $(-3,6)$ and the coefficients of the line $3x - y - 5 = 0$ into the distance formula.
$d = \frac{|3(-3) - 1(6) - 5|}{\sqrt{3^2 + (-1)^2}}$

STEP 5

Calculate the numerator of the distance formula.
$Numerator = |3(-3) - 6 - 5| = |-9 - 6 - 5| = |-20| = 20$

STEP 6

Calculate the denominator of the distance formula.
$Denominator = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$

STEP 7

Now, calculate the distance $d$ , which is the radius of the circle.
$d = \frac{20}{\sqrt{10}} = \frac{20}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{20\sqrt{10}}{10} = 2\sqrt{10}$

STEP 8

Now that we have the radius of the circle, we can write the equation of the circle using the general form.

STEP 9

Substitute the center of the circle $(-3,6)$ and the radius $2\sqrt{10}$ into the general equation of a circle.
$(x - (-3))^2 + (y - 6)^2 = (2\sqrt{10})^2$

STEP 10

Simplify the equation of the circle.
$(x + 3)^2 + (y - 6)^2 = 4 \cdot 10$

STEP 11

Further simplify the equation of the circle by expanding the right side.
$(x + 3)^2 + (y - 6)^2 = 40$

STEP 12

The equation of the circle in standard form is:
$(x + 3)^2 + (y - 6)^2 = 40$
This is the final equation of the circle with center $(-3,6)$ that is tangent to the line $3x - y - 5 = 0$ .

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Question Find the equation of a circle with center at $(-3, 6)$ given that the line $3x - y - 5 = 0$ is tangent to the circle.

Studdy Solution

STEP 1

STEP 2

First, we need to find the perpendicular distance from the center of the circle to the tangent line. This distance will be the radius of the circle.

STEP 3

The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula:
$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

STEP 4

Plug in the values for the center of the circle $(-3,6)$ and the coefficients of the line $3x - y - 5 = 0$ into the distance formula.
$d = \frac{|3(-3) - 1(6) - 5|}{\sqrt{3^2 + (-1)^2}}$

STEP 5

Calculate the numerator of the distance formula.
$Numerator = |3(-3) - 6 - 5| = |-9 - 6 - 5| = |-20| = 20$

STEP 6

Calculate the denominator of the distance formula.
$Denominator = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$

STEP 7

Now, calculate the distance $d$ , which is the radius of the circle.
$d = \frac{20}{\sqrt{10}} = \frac{20}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{20\sqrt{10}}{10} = 2\sqrt{10}$

STEP 8

Now that we have the radius of the circle, we can write the equation of the circle using the general form.

STEP 9

Substitute the center of the circle $(-3,6)$ and the radius $2\sqrt{10}$ into the general equation of a circle.
$(x - (-3))^2 + (y - 6)^2 = (2\sqrt{10})^2$

STEP 10

Simplify the equation of the circle.
$(x + 3)^2 + (y - 6)^2 = 4 \cdot 10$

STEP 11

Further simplify the equation of the circle by expanding the right side.
$(x + 3)^2 + (y - 6)^2 = 40$

STEP 12

The equation of the circle in standard form is:
$(x + 3)^2 + (y - 6)^2 = 40$
This is the final equation of the circle with center $(-3,6)$ that is tangent to the line $3x - y - 5 = 0$ .

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Question Find the equation of a circle with center at (−3,6)(-3, 6)(−3,6) given that the line 3x−y−5=03x - y - 5 = 03x−y−5=0 is tangent to the circle.

Studdy Solution

STEP 1

STEP 2

First, we need to find the perpendicular distance from the center of the circle to the tangent line. This distance will be the radius of the circle.

STEP 3

The distance ddd from a point (x0,y0)(x_0, y_0)(x0​,y0​) to a line Ax+By+C=0Ax + By + C = 0Ax+By+C=0 is given by the formula:d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}d=A2+B2​∣Ax0​+By0​+C∣​

STEP 4

Plug in the values for the center of the circle (−3,6)(-3,6)(−3,6) and the coefficients of the line 3x−y−5=03x - y - 5 = 03x−y−5=0 into the distance formula.d=∣3(−3)−1(6)−5∣32+(−1)2d = \frac{|3(-3) - 1(6) - 5|}{\sqrt{3^2 + (-1)^2}}d=32+(−1)2​∣3(−3)−1(6)−5∣​

STEP 5

Calculate the numerator of the distance formula.Numerator=∣3(−3)−6−5∣=∣−9−6−5∣=∣−20∣=20Numerator = |3(-3) - 6 - 5| = |-9 - 6 - 5| = |-20| = 20Numerator=∣3(−3)−6−5∣=∣−9−6−5∣=∣−20∣=20

STEP 6

Calculate the denominator of the distance formula.Denominator=32+(−1)2=9+1=10Denominator = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}Denominator=32+(−1)2​=9+1​=10​

STEP 7

Now, calculate the distance ddd, which is the radius of the circle.d=2010=2010⋅1010=201010=210d = \frac{20}{\sqrt{10}} = \frac{20}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{20\sqrt{10}}{10} = 2\sqrt{10}d=10​20​=10​20​⋅10​10​​=102010​​=210​

STEP 8

Now that we have the radius of the circle, we can write the equation of the circle using the general form.

STEP 9

Substitute the center of the circle (−3,6)(-3,6)(−3,6) and the radius 2102\sqrt{10}210​ into the general equation of a circle.(x−(−3))2+(y−6)2=(210)2 (x - (-3))^2 + (y - 6)^2 = (2\sqrt{10})^2 (x−(−3))2+(y−6)2=(210​)2

STEP 10

Simplify the equation of the circle.(x+3)2+(y−6)2=4⋅10 (x + 3)^2 + (y - 6)^2 = 4 \cdot 10 (x+3)2+(y−6)2=4⋅10

STEP 11

Further simplify the equation of the circle by expanding the right side.(x+3)2+(y−6)2=40 (x + 3)^2 + (y - 6)^2 = 40 (x+3)2+(y−6)2=40

STEP 12

The equation of the circle in standard form is:(x+3)2+(y−6)2=40 (x + 3)^2 + (y - 6)^2 = 40 (x+3)2+(y−6)2=40This is the final equation of the circle with center (−3,6)(-3,6)(−3,6) that is tangent to the line 3x−y−5=03x - y - 5 = 03x−y−5=0.

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Question Find the equation of a circle with center at $(-3, 6)$ given that the line $3x - y - 5 = 0$ is tangent to the circle.

The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula:
$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Plug in the values for the center of the circle $(-3,6)$ and the coefficients of the line $3x - y - 5 = 0$ into the distance formula.
$d = \frac{|3(-3) - 1(6) - 5|}{\sqrt{3^2 + (-1)^2}}$

Calculate the numerator of the distance formula.
$Numerator = |3(-3) - 6 - 5| = |-9 - 6 - 5| = |-20| = 20$

Calculate the denominator of the distance formula.
$Denominator = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$

Now, calculate the distance $d$ , which is the radius of the circle.
$d = \frac{20}{\sqrt{10}} = \frac{20}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{20\sqrt{10}}{10} = 2\sqrt{10}$

Substitute the center of the circle $(-3,6)$ and the radius $2\sqrt{10}$ into the general equation of a circle.
$(x - (-3))^2 + (y - 6)^2 = (2\sqrt{10})^2$

Simplify the equation of the circle.
$(x + 3)^2 + (y - 6)^2 = 4 \cdot 10$

Further simplify the equation of the circle by expanding the right side.
$(x + 3)^2 + (y - 6)^2 = 40$

The equation of the circle in standard form is:
$(x + 3)^2 + (y - 6)^2 = 40$
This is the final equation of the circle with center $(-3,6)$ that is tangent to the line $3x - y - 5 = 0$ .