Question

Write the equation of the circle described.
b. diameter with endpoints (2,5) and (-10, 7)

Answer 1

Answer :

• (x + 4)^2 + (y - 6)^2 = 37

Explanation :

firstly,find the centre of the circle using the midpoint formula

Midpoint = (x1 +x2)/2 ,(y1 +y2)/2

Midpoint = (2 -10)/2,(5 +7)/2

• Midpoint = (-4,6)

Now, using the distance formula,find the radius of the circle

distance = √((x2 - x1)^2 + (y2 - y1)^2)

radius = √((2 -(-4))^2 + (5 - 6)^2)

radius = √(36 + 1 )

• radius = √37

now, we can write the equation of the circle which is given by,

• (x - h)^2 + (y - k)^2 = r^2

where,

• (h,k) = centre of the circle
• r = radius,

thus, the equation of the circle would be

• (x - (-4)^2 + (y - 6)^2 = (√37)^2
• (x + 4)^2 + (y - 6)^2 = 37

Answer 2

Answer:

$(x+4)^2+(y-6)^2=37$

Step-by-step explanation:

To find the equation of a circle given its diameter with endpoints (2, 5) and (-10, 7), we first need to find the center of the circle and the radius.

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Center of the circle

The center of the circle is the midpoint of the diameter. Therefore, to find the center we can use the midpoint formula:

$\boxed{\begin{array}{l}\underline{\sf Midpoint \;formula}\\\\M(x,y) =\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\\\\\textsf{where (x_1,y_1) and (x_2,y_2) are the endpoints.}\\\end{array}}$

In this case:

• M = center (h, k)
• (x₁, y₁) = (2, 5)
• (x₂, y₂) = (-10, 7)

Substitute the values into the midpoint formula:

$\textsf{Center (h,k) }=\left(\dfrac{-10+2}{2},\dfrac{7+5}{2}\right)\\\\\\\textsf{Center (h,k) }=\left(\dfrac{-8}{2},\dfrac{12}{2}\right)\\\\\\\textsf{Center (h,k) }=\left(-4,6\right)$

So, the center of the circle is (-4, 6).

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Radius of the circle

The radius of the circle is the distance between the center and one of the endpoints. Therefore, to find the radius of the circle, we can substitute the coordinates of the center and one of the diameter endpoints into the distance formula:

$\boxed{\begin{array}{l}\underline{\sf Distance \;Formula}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\;d\;\textsf{is the distance between two points.} \\\phantom{ww}\bullet\;\;\textsf{ (x_1,y_1) and (x_2,y_2) are the two points.}\end{array}}$

In this case:

• d = radius (r)
• (x₁, y₁) = center = (-4, 6)
• (x₂, y₂) = (-10, 7)

Substitute the values into the distance formula:

$r=\sqrt{(-10-(-4))^2+(7-6)^2}\\\\r=\sqrt{(-6)^2+(1)^2}\\\\r=\sqrt{36+1}\\\\r=\sqrt{37}$

Therefore, the radius of the circle is $\sqrt{37}$.

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Equation of the circle

The standard form of the equation of a circle is:

$\boxed{\begin{array}{l}\underline{\textsf{General equation of a circle}}\\\\(x-h)^2+(y-k)^2=r^2\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{ (h, k) is the center.}\\\phantom{ww}\bullet\;\textsf{ r is the radius.}\end{array}}$

Substitute the found center (-4, 6) and radius $\sqrt{37}$ into the formula:

$(x-(-4))^2+(y-6)^2=\left(\sqrt{37}\right)^2\\\\\\(x+4)^2+(y-6)^2=37$

Therefore, the equation of the circle is:

$\Large\boxed{\boxed{(x+4)^2+(y-6)^2=37}}$