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

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Answer 1 · 2017-02-15T14:25:39

$(x+2)^2+(y-3)^2=(145/sqrt841)^2$ . See the tangent-inclusive Socratic graph.

The length of the perpendicular from the center $(-2. 3)$ to the

tangent line 20x-21y-42=0#

$=|20(-2)-21(3)-42|/sqrt(20^2+21^2)=145/sqrt841$

= radius of the circle.

So, the equation is

$(x+2)^2+(y-3)^2=(145/sqrt841)^2$

graph{(x-21/20y-2.1)((x+2)^2+(y-3)^2-(145^2/841))=0 [-20, 10, -5, 10]}

How do you write the standard form equation for the circle whose center is at (-2, 3)(-2, 3) and that is tangent to the line 20x - 21y - 42 = 020x - 21y - 42 = 0?