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PilotLPTM [1.2K]
3 years ago
11

Write the equation of circle b with center B(-2,3) that passes through (1,2)

Mathematics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

(x+2)^2 + (y-3)^2 = 10

Step-by-step explanation:

The standard equation for circle is

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

where point (a,b) is coordinate of center of circle and r is the radius.

______________________________________________________

Given

center of circle =((-2,3)

let r be the radius of circle

Plugging in this value of center in standard equation for circle given above we have

(x-a)^2 + (y-b)^2 = r^2    \ substitute (a,b) \ with (-2,3) \\=>(x-(-2))^2 + (y-3)^2 = r^2  \\=>(x+2)^2 + (y-3)^2 = r^2    (1)

Given that point (1,2 ) passes through circle. Hence this point will satisfy the above equation of circle.

Plugging in the point (1,2 )  in equation 1 we have

\\=>(x+2)^2 + (y-3)^2 = r^2    \\=> (1+2)^2 + (2-3)^2 = r^2\\=> 3^2 + (-1)^2 = r^2\\=> 9 + 1 = r^2\\=> 10 = r^2\\=>  r^2  = 10\\

now we have value of r^2 = 10, substituting this in equation 1 we have

Thus complete equation of circle is =>(x+2)^2 + (y-3)^2 = r^2\\=>(x+2)^2 + (y-3)^2 = 10

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