a. Setting the right side to 0 and expanding each equation gives:

Equation [1]:

(

x+ 2)^{2}+ (y− 3)^{2}= 25(

x+ 2)^{2}+ (y− 3)^{2}− 25 = 0

x^{2}+ 4x+ 4 +y^{2}− 6y+ 9 − 25 = 0

x^{2}+ 4x+y^{2}− 6y− 12 = 0 [3]

Equation [2]:

(

x− 1)^{2}+ (y+ 4)^{2}= 16(

x− 1)^{2}+ (y+ 4)^{2}− 16 = 0

x^{2}− 2x+ 1 +y^{2}+ 8y+ 16 − 16 = 0

x^{2}− 2x+y^{2}+ 8y+ 1 = 0 [4]

Solving the above 2 results simultaneously, we subtract Row [4] from Row [3] and this gives:

6

x− 14y− 13 = 0

Solving for *y* gives:

`y=3/7x-13/14`

This means the intersection points are on the line

`y=3/7x-13/14`

b. Solve one of the circle equations for *y* using:

`y=(-b+-sqrt(b^2-4ac))/(2a)`

`y^2-6y+(x^2+4x-12)=0`

This gives:

`y=3+-sqrt(21-x^2-4x`

c. Substitute the positive case into the LHS of

`y=3/7x-13/14`, gives us:

`3+sqrt(21-x^2-4x)=3/7x-13/14`

d. Solve for *x*:

`sqrt(21-x^2-4x)` `=3/7x-13/14-3` `=3/7x-55/14`

Square both sides:

`21-x^2-4x` ` =(3/7x-55/14)^2` `=3025/196-165/49x+9/49x^2`

Re-writing for convenience:

`21-x^2-4x` `=3025/196-165/49x+9/49x^2`

Moving everything to the right hand side:

`0=9/49x^2+x^2-165/49x` `+4x` `+3025/196` `-21`

Simplifying:

`0=58/49x^2+31/49x-1091/196`

This gives us a quadratic in *x*

`58/49x^2+31/49x-1091/196=0`

or more simply

`58x^2+31x-272 3/4=0`

`x=-31/116+7/116sqrt1311` `=1.9177`

`x=-31/116-7/116sqrt1311` `=-2.4522`

e. Substitute these two *x*-values into

`y=3/7x-13/14`:

`[3/7x-13/14]_(x=1.9177)=-0.1067`

`[3/7x-13/14]_(x=-2.4522)=-1.9795`

So the points of intersection are (1.9177, −0.1067) and (−2.4522, −1.9795).

[We could have substituted the *x*-values into either circle equation and solved for *y*, but what I have done is easier.]

We can see that the circles, the line `y=3/7x-13/14` and the intersection points are all correct when we draw the graph:

As a comment, this question could be solved very quickly, and easily, using a computer graphics program. We would just need to zoom in on the intersection points until we obtained the required precision.

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