The equation of the circle which passing thro | vedclass

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

Question

(a) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

Since it passes through $(2a,\;0)$, so

$4{a^2} + 4ag + c = 0$.&hellip;.$(i)$

Vedclass · Accepted Answer

${x^2} + {y^2} - 2ax = 0$

Vedclass · Answer

(a) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

Since it passes through $(2a,\;0)$, so

$4{a^2} + 4ag + c = 0$.&hellip;.$(i)$

Also its radical axis with ${x^2} + {y^2} = {a^2}$ is

$2gx + 2fy + c + a \equiv x - \frac{a}{2} = 0$

$ \Rightarrow \frac{{2g}}{1} = \frac{{c + a}}{{ - a/2}}$ and $f = 0$

or $ag + c + a = 0$.&hellip;.$(ii)$

From $(i)$ and $(ii),$ we get $g$ and $c$.

Hence equation is ${x^2} + {y^2} - 2ax = 0$.

The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be

Similar Questions

The number of common tangents of the circles given by $x^2 +y^2 - 8x - 2y + 1 = 0$ and $x^2 + y^2 + 6x + 8y = 0$ is

The two circles ${x^2} + {y^2} - 2x + 6y + 6 = 0$ and ${x^2} + {y^2} - 5x + 6y + 15 = 0$

If the circles ${x^2} + {y^2} + 2ax + cy + a = 0$ and ${x^2} + {y^2} - 3ax + dy - 1 = 0$ intersect in two distinct points $P$ and $Q$ then the line $5x + by - a = 0$ passes through $P$ and $Q$ for

If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4 x-4 y+4=0$ intersect at exactly two distinct points, then

Let the equation $x^{2}+y^{2}+p x+(1-p) y+5=0$ represent circles of varying radius $\mathrm{r} \in(0,5]$. Then the number of elements in the set $S=\left\{q: q=p^{2}\right.$ and $\mathrm{q}$ is an integer $\}$ is ..... .