How do you find the equation of a circle center is at #(-2, 3)# and that is tangent to the line #20x - 21y - 42 = 0#?

Question

5472 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-31T11:24:08

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

or,

# x^2+y^2+4x-6y-12=0.#

Let us name the Centre #C,# Radius #r# and, the Tgt. Line #t.#

#:. C=C(-2,3), and, t :20x-21y-42=0.#

From Geometry, we know that, the #bot-#dist. from #C# to #t# is #r#.

Recall that the #bot-#dist. from pt. #(h,k)# to line #l : ax+by+c=0#

is #|ah+bk+c|/sqrt(a^2+b^2)#.

#:. r=|20(-2)-21(3)-42|/sqrt(20^2+(-21)^2)=145/29=5.#

Hence follows the Eqn. of the Circle, #(x+2)^2+(y-3)^2=5^2,# or,

# x^2+y^2+4x-6y-12=0.#

Enjoy Maths.!