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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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You will be using the equation y = mx + b

The only variables you need to find are "m" and "b"

For the equation of a line to be perpendicular to the given line, the slope will have to be the opposite of the given slope. Or you have to flip the sign and the number of the given slope to get your new slope.

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