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

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Answer 1 · 2017-03-31T11:24:08

${(x + 2)}^{2} + {(y - 3)}^{2} = 5^{2},$ $(x+2)^2+(y-3)^2=5^2,$

or,

$x^{2} + y^{2} + 4 x - 6 y - 12 = 0 .$ $x^2+y^2+4x-6y-12=0.$

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

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

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)(-2, 3) and that is tangent to the line 20x - 21y - 42 = 020x−21y−42=020x - 21y - 42 = 0?