Given the circle $(x-4)^2 + (y+6)^2 =4$ , determine all values of the real number c if the direction $3x+2y=c$ is the tangent of the circle?

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Answer 1 · 2016-09-11T11:26:07

$c=+-sqrt13$ .

Observe that the Centre $C$ of the circle is

$C(4,-6)", and, radius r=2"$ .

If the given line $L : 3x+2y-c=0$ is a tgt. to the circle, then, from

Geometry, we know that,

$"The" bot"-distance from "C" to "L=r$ .

Recall that

#"The" bot"-distance from "(x_1,y_1)" to line" : lx+my+n=0 "is"

$|lx_1+my_1+n|/sqrt(l^2+m^2)$ .

Hence, $|3*4+2(-6)-c|/sqrt(9+4)=2, i.e., |c|=2sqrt13$ .

$:. c=+-sqrt13$ .

Given the circle (x-4)^2 + (y+6)^2 =4(x-4)^2 + (y+6)^2 =4, determine all values of the real number c if the direction 3x+2y=c3x+2y=c is the tangent of the circle?