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write the standard form equation for the circle whose center is at (-2, 3) and that is tangent to the line 20x - 21y - 42 = 0.

a). x^2 - 4x + y^2 + 6y - 12 = 0
b). x^2 + 4x + y^2 - 6y - 12 = 0
c). x^2 + 4x + y^2 - 6y - 25 = 0

➷ Use the geometry rule:

(|(20(-2) - 21(3) - 42|) / square root of 20^2 + (-21)^2

Now you get:

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

This can also be written as:

x^2 + 4x + y^2 - 6y - 12.

It would be option B

➶ Hope This Helps You!

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➶ Have A Great Day ^-^

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