For the circles (x-a)^2+y^2=a^2 and x^2+(y-a) | vedclass

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(D) Given circles are $C_1: (x-a)^2+y^2=a^2$ and $C_2: x^2+(y-a)^2=a^2$. Expanding these,we get $x^2-2ax+a^2+y^2=a^2 \Rightarrow x^2+y^2-2ax=0$ and $x

Vedclass · Accepted Answer

common chord is at a distance of $(\sqrt{2} a)$ units from the centres of given circles

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(D) Given circles are $C_1: (x-a)^2+y^2=a^2$ and $C_2: x^2+(y-a)^2=a^2$. Expanding these,we get $x^2-2ax+a^2+y^2=a^2 \Rightarrow x^2+y^2-2ax=0$ and $x^2+y^2-2ay=0$. The equation of the common chord is $C_1 - C_2 = 0$,which gives $-2ax + 2ay = 0 \Rightarrow x-y=0$. Substituting $y=x$ into $x^2+y^2-2ax=0$,we get $2x^2-2ax=0 \Rightarrow 2x(x-a)=0$. Thus,the intersection points are $(0,0)$ and $(a,a)$. The mid-point of the chord is $(\frac{0+a}{2}, \frac{0+a}{2}) = (\frac{a}{2}, \frac{a}{2})$. The length of the chord is $\sqrt{(a-0)^2+(a-0)^2} = \sqrt{2a^2} = \sqrt{2}a$. The distance from the center $(a,0)$ to the line $x-y=0$ is $d = \frac{|a-0|}{\sqrt{1^2+(-1)^2}} = \frac{a}{\sqrt{2}}$. Since $\frac{a}{\sqrt{2}} 
eq \sqrt{2}a$,option $D$ is not true.

For the circles $(x-a)^2+y^2=a^2$ and $x^2+(y-a)^2=a^2$,where $a>0$,which one of the following is not true?

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