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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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?

Similar Questions

The equation of the circle having the common chord of the circles $x^2+y^2-8x=0$ and $x^2+y^2-9=0$ as its diameter is

Let the circle $C_1: x^2+y^2-2(x+y)+1=0$ and $C_2$ be a circle having centre at $(-1,0)$ and radius $2$. If the line of the common chord of $C_1$ and $C_2$ intersects the $y$-axis at the point $P$,then the square of the distance of $P$ from the centre of $C_1$ is:

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then $(4g-3)(f-2)=$

Given the circles $x^2 + y^2 - 4x - 5 = 0$ and $x^2 + y^2 + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha, \beta)$ such that the lengths of the tangents from $P$ to both circles are equal. Then:

If the circles $x^2+y^2+5kx+2y+k=0$ and $2x^2+2y^2+2kx+3y-1=0$,$k \in R$ intersect at points $P$ and $Q$,then the line $4x+5y-k=0$ passes through $P$ and $Q$ for

Vedclass Test Series

Exam Paper Generator

Online Exam Module