a. Prove that the circles x^2 + y^2 + 2ax + c^2 = 0 and x^2 + y^2 + 2by + c^2 = 0 touch each other if 1a^2 + 1b^2 = 1c^2 b. Evaluate: x x x

a. Proving the circles touch if $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$

Step 1: Rewrite the circle equations in standard form

The given circles are:

1. $x^2+y^2+2ax+c^2=0$
2. $x^2+y^2+2by+c^2=0$

Rewrite them in standard form $(x-h{)}^2+(y-k{)}^2=r^2$ by completing the square.

For the first circle: $$x^2+2ax+y^2=-c^2$$ $$(x^2+2ax+a^2)+y^2=a^2-c^2$$ $$(x+a{)}^2+y^2=a^2-c^2$$ So, the center is $(-a,0)$ and the radius is $r_1=\sqrt{a^2-c^2}$.

For the second circle: $$x^2+y^2+2by=-c^2$$ $$x^2+(y^2+2by+b^2)=b^2-c^2$$ $$x^2+(y+b{)}^2=b^2-c^2$$ So, the center is $(0,-b)$ and the radius is $r_2=\sqrt{b^2-c^2}$.

Step 2: Distance between centers

The distance $d$ between the centers $(-a,0)$ and $(0,-b)$ is: $$d=\sqrt{(0-(-a){)}^2+(-b-0{)}^2}=\sqrt{a^2+b^2}$$

Step 3: Condition for touching circles

Two circles touch if the distance between their centers equals the sum or difference of their radii. Here, we assume they touch externally (sum of radii): $$d=r_1+r_2$$ $$\sqrt{a^2+b^2}=\sqrt{a^2-c^2}+\sqrt{b^2-c^2}$$

Square both sides: $$a^2+b^2=(a^2-c^2)+(b^2-c^2)+2\sqrt{(a^2-c^2)(b^2-c^2)}$$ $$a^2+b^2=a^2+b^2-2c^2+2\sqrt{(a^2-c^2)(b^2-c^2)}$$ $$0=-2c^2+2\sqrt{(a^2-c^2)(b^2-c^2)}$$ $$2c^2=2\sqrt{(a^2-c^2)(b^2-c^2)}$$ $$c^2=\sqrt{(a^2-c^2)(b^2-c^2)}$$ Square again: $$c^4=(a^2-c^2)(b^2-c^2)$$ $$c^4=a^2{b}^2-a^2{c}^2-b^2{c}^2+c^4$$ $$0=a^2{b}^2-a^2{c}^2-b^2{c}^2$$ $$a^2{b}^2=c^2(a^2+b^2)$$ Divide both sides by $a^2{b}^2{c}^2$: $$\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$$

This matches the given condition. Thus, the circles touch if $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$.

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b. Evaluate ${\lim }_{x\to \theta }\frac{\sin x-\sin \theta }{x-\theta }$

Step 1: Rewrite the expression

The limit is of the form $\frac{0}{0}$ when $x\to \theta$, so apply L'Hôpital's Rule (differentiate numerator and denominator): $$\underset{x\to \theta }{\lim }\frac{\sin x-\sin \theta }{x-\theta }=\underset{x\to \theta }{\lim }\frac{\cos x}{1}=\cos \theta$$

Step 2: Alternative approach using trigonometric identity

Use the identity $\sin x-\sin \theta =2\cos \left(\frac{x+\theta }{2}\right)\sin \left(\frac{x-\theta }{2}\right)$: $$\frac{\sin x-\sin \theta }{x-\theta }=\frac{2\cos \left(\frac{x+\theta }{2}\right)\sin \left(\frac{x-\theta }{2}\right)}{x-\theta }$$ As $x\to \theta$, $\sin \left(\frac{x-\theta }{2}\right)\approx \frac{x-\theta }{2}$ (small-angle approximation): $$\underset{x\to \theta }{\lim }\frac{2\cos \left(\frac{x+\theta }{2}\right)\cdot \frac{x-\theta }{2}}{x-\theta }=\underset{x\to \theta }{\lim }\cos \left(\frac{x+\theta }{2}\right)=\cos \theta$$

Final result:

$$\boxed{\cos \theta }$$