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Circle 1:

$$x^2 + y^2 +6x + 2y +6 = 0$$

Circle 2:

$$x^2 + y^2 + 8x + y + 10 = 0$$

My attampt:

From circle 1 and 2, I found

$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got

$$5x^2 + 26x + 30 = 0$$

here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?

First, obtain the equations of the intersection points below for both $x$ and $y$,

$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$

It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,

$$x_1+x_2=-frac265,>>>x_1x_2=6$$

$$y_1+y_2=-frac125,>>>y_1y_2=-frac 85 $$

Thus, the center of the circle is $(-13/5, -6/5)$ and its diameter squared is,

$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$

$$= left( frac265 right)^2 -4cdot 6 + left( frac125right)^2 + 4cdot frac 85 = frac765$$

The equation of the circle is

$$left( x+frac135 right)^2 + left( y +frac65right)^2 = frac195$$

You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 Rightarrow x_1=frac-13-sqrt195,x_2=frac-13+sqrt195\
y_1=frac-6-2sqrt195, y_2=frac-6+2sqrt195$$
The center of the new circle:
$$fracx_1+x_22=-frac135,fracy_1+y_22=-frac65$$
The diameter of the new circle:
$$d=sqrt(x_1-x_2)^2+(y_1-y_2)^2=sqrtfrac4cdot 1925+frac16cdot 1925=sqrtfrac765 Rightarrow \
r=frac12d=sqrtfrac764cdot 5=sqrtfrac195$$
Thus:
$$left(x+frac135right)^2+left(y+frac65right)^2=frac195.$$

You can certainly keep going the way you are: solve the quadratic equation for $x$ and substitute back into the equation of the radical axis (or either circle equation) to get the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of the chord. An equation of the circle with that diameter can be written down directly: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Rearrange this into whatever form is required.

However, there’s another way to solve this that requires less work, I think. The circles that pass through the two intersection points form a one-parameter family with equations that can be expressed as an affine combination of the equations of the two circles, to wit, $$(1-lambda)(x^2+y^2+6x+2y+6)+lambda(x^2+y^2+8x+y+10) = 0$$ or $$x^2+y^2+(2lambda+6)x+(2-lambda)y+(4lambda+6)=0.tag*$$ By inspection, the coordinates of the center of this circle are $(-lambda-3,lambda/2-1)$ and we want it to lie on the radical axis $y=2x+4$. Substitute the coordinates of the center into this equation and solve the resulting linear equation for $lambda$, then plug that value into equation (*).