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4vir4ik [10]
3 years ago
12

Please read below. Thank you.

Mathematics
2 answers:
mel-nik [20]3 years ago
6 0

Answer:

The equation of the circle is given by:

(x-a)^2+(y-b)^2=r^2

where:

(a,b) is the center  of the circle

given that the center of our circle is (2,3) with the radius of 5, the equation will be:

(x+2)^2+(y+3)^2=5^2

expanding the above we get:

x^2+4x+4+y^2+6y+9=25

this can be simplified to be:

x^2+4x+y^2+6y=25-13

x^2+y^2+4x+6y=12

iren2701 [21]3 years ago
3 0

Answer:

\sf\longrightarrow \boxed{\sf x^2+y^2-4x-6y-12=0}

Step-by-step explanation:

Here we are given the radius of circle as <u>5</u><u>c</u><u>m</u> and the centre of the circle is (2,3) . We need to find the equation of the circle. Here we can yse the <u>Standard</u><u> equation</u><u> of</u><u> </u><u>circle</u> to find the equation .

<u>Standard</u><u> equation</u><u> of</u><u> </u><u>circle</u><u> </u><u>:</u><u>-</u>

\sf\implies \green{ (x - h )^2+(y-k)^2 = r^2 }

  • where (h,k) is the centre and r is radius .

Substitute the respective values ,

\sf\longrightarrow ( x - 2 )^2 + ( y - 3)^2 = 5^2

Simplify the whole square ,

\sf\longrightarrow x^2 + 4 -4x + y^2+9-6y = 25

Rearrange and add the constants ,

\sf\longrightarrow x^2 + y^2 -4x -6y +13 = 25

Subtract 25 on both sides ,

\sf\longrightarrow x^2 +y^2-4x-6y+13-25=0

Simplify ,

\sf\longrightarrow \boxed{\blue{\sf x^2+y^2-4x-6y-12=0}}

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