How do you write an equation for a circle with tangent to 3x - 4y + 3 = 0 at (-1,0) radius 7?

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Answer 1 · 2016-09-13T20:32:37

Thus, there are two circles satisfying the given cond.

$(1) : (x+26/5)^2+(y-28/5)^2=49$ .

$(2) : (x-16/5)^2+(y+28/5)^2=49$ .

Suppose that $C(h,k)$ is the Centre of the circle.

Since, the pt. of contact $P(-1,0)$ is on the circle,

$CP=radius=7$

$:. (h+1)^2+k^2=49........................(1)$ .

Also, line $CP bot" the tgt. : "3x-4y+3=0$

As slope of tgt. is $3/4, :. "slope of "CP=-4/3$ .

But, slope of $CP" is "k/(h+1) :. k/(h+1)=-4/3$ .

$:. k/4=(h+1)/-3=mu," say, so that, "k=4mu, h+1=-3mu$ .

Sub.ing in $(1)$ , we have, $9mu^2+16mu^2=49 rArr mu=+-7/5$

$Case 1 : mu=+7/5$

$:. k=4mu=28/5, h=-1-3mu=-1-21/5=-26/5$

So, the Centre is $C(h,k)=C(-26/5,28/5), radius=7$ . Hence,

the eqn. of the circle, is, $(x+26/5)^2+(y-28/5)^2=49$ .

$Case 2 : mu=-7/5$

In this case, the Centre is $C(16/5,-28/5)$ , & the eqn. of circle is

$(x-16/5)^2+(y+28/5)^2=49$ .

Thus, there are two circles satisfying the given cond.

$(1) : (x+26/5)^2+(y-28/5)^2=49$ .

$(2) : (x-16/5)^2+(y+28/5)^2=49$ .

Enjoy Maths.!