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]}