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

1 Answer
Sep 13, 2016

Answer:

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#.

Explanation:

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.!