Query 1 :

Write the equation of a circle in commonplace kind whose heart is (3, 2) and radius is 6.

**Query 2 :**

Write the equation of a circle in commonplace kind whose heart is (0, -3) and radius is 7.

**Query 3 :**

If two finish factors of the diameter of a circle are (-1, -2) and (11, 14), discover the equation of the circle in commonplace kind.Â

**Query 4 :**

Write the next basic type of equation of the circle in commonplace kind. And discover the middle and radius.Â

x^{2}Â + y^{2Â }– 4x + 6y – 12Â =Â 0

**Query 5 :**

The purpose (1 ,2) is on a circle whose heart is (5, -1).Â Write the equation of the circle in commonplace equation.Â

Query 6 :

Discover the equation of the circle proven under in commonplace kind.

**1. Reply :**

Equation of a circle in commonplace kind :

(x – h)^{2}Â + (y – ok)^{2}Â =Â r^{2}

Substitute (h, ok) = (3, 2) and r = 6.

(x – 3)^{2}Â + (y – 2)^{2}Â =Â 6^{2}

(x – 3)^{2} + (y – 2)^{2}Â =Â 36

**2. Reply :**

Equation of a circle in commonplace kind :

(x – h)^{2}Â + (y – ok)^{2}Â =Â r^{2}

Substitute (h, ok) = (0, -3) and r = 7.

(x – 0)^{2}Â + [y – (-3)^{2}]Â =Â 7^{2}

x^{2}Â + (y + 3)^{2}Â =Â 7^{2}

**3. Reply :**

Utilizing mid level method, discover the mid level of the diameter.Â Â

Mid level of (-1, -2) and (11, 14) :Â

=Â [(-1 + 11)/2, (-2 + 14)/2]

=Â (10/2, 12/2)

=Â (5, 6)

Mid level of the diameter is the middle of the circle.Â

So, the middle of the circle is (5, 6).Â

Utilizing distance method to seek out the space between the middle and one of many finish factors of the diameter.

Distance between (5, 6) and (11, 14) :

=Â âˆš[(11 – 5)^{2}Â + (14 – 6)^{2}]

=Â âˆš[6^{2}Â + 8^{2}]

=Â Â âˆš[36Â + 64]

=Â Â âˆš100

=Â 10

Radius of the circle is 10 items.Â

Equation of the circle in commonplace kind :Â

(x – h)^{2}Â + (y – ok)^{2}Â =Â r^{2}

Substitute (h, ok) = (5, 6) and r = 10.

(x – 5)^{2}Â + (y – 6)^{2}Â =Â 10^{2}

**4. Reply :**

x^{2}Â + y^{2Â }– 4x + 6y – 12Â =Â 0

Write x-terms and y-terms collectively.Â

x^{2}Â – 4x + y^{2}Â + 6y – 12Â =Â 0

Finishing the sq..Â

x^{2}Â – 2(x)(2) + 2^{2}Â –Â 2^{2Â }+ y^{2}Â + 2(y)(3) + 3^{2}–Â 3^{2}Â – 12Â =Â 0

(xÂ – 2)^{2}Â –Â 2^{2Â }+ (y + 3)^{2Â }–Â 3^{2}Â – 12Â =Â 0

(xÂ – 2)^{2}Â – 4^{Â }+ (y + 3)^{2Â }– 9Â – 12Â =Â 0

(xÂ – 2)^{2Â }+ (y + 3)^{2Â }–Â 25Â =Â 0

Add 25 to every facet.

(xÂ – 2)^{2Â }+ (y + 3)^{2}Â =Â 25

(xÂ – 2)^{2Â }+ (y + 3)^{2}Â =Â 5^{2}

The above equation of the circle is in commonplace kind.Â Â

Heart (h, ok)Â =Â (2, -3)

RadiusÂ =Â 5

**5. Reply :**

To seek out the usual equation of the circle, we have to know the middle and radius.Â The middle is already given and we have to discover the radius.

Utilizing distance method, we’ve got

RadiusÂ =Â Â âˆš[(5 – 1)^{2}Â + (-1 – 2)^{2}]

RadiusÂ =Â Â âˆš[4^{2}Â + (-3)^{2}]

RadiusÂ =Â Â âˆš[16 + 9]

RadiusÂ =Â Â âˆš25

RadiusÂ =Â 5

Equation of a circle in commonplace kind :Â

(x – h)^{2}Â + (y – ok)^{2}Â =Â r^{2}

Substitute (h, ok) = (5, -1) and r = 5.

(x – 5)^{2}Â + [y – (-1)^{2}]Â =Â 5^{2}

Simplify.Â

(x – 5)^{2}Â + (y + 1)^{2}Â =Â Â 5^{2}

**6. Reply :**

Within the given circle, measure the size of the radius.

The circle above has the middle (5, 2) and radius 4 items.

Equation of the circle in commonplace kind :

(x – h)^{2} + (y – ok)^{2} = r^{2}

Substitute (h, ok) = (5, 2) and r = 4.

(x – 5)^{2}Â + (y – 2)^{2}Â = 4^{2}

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