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

5748 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.!

How do you find the equation of a circle center is at (-2, 3)(-2, 3) and that is tangent to the line 20x - 21y - 42 = 020x - 21y - 42 = 0?