Equation of radical axis of circles {x^2} + {y^2} - 3x - 4y + 5 = 0 , 2{x^2} + 2{y^2} - 10x - 12y +

English

Gujarati

Hindi

10-1.Circle and System of Circles

medium

Equation of radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$, $2{x^2} + 2{y^2} - 10x$$ - 12y + 12 = 0$ is

A

$2x + 2y - 1 = 0$

B

$2x + 2y + 1 = 0$

C

$x + y + 7 = 0$

D

$x + y - 7 = 0$

Solution

(a) Circles are,

${S_1} = {x^2} + {y^2} – 3x – 4y + 5 = 0$ …..$(i)$

and ${S_2} = 2{x^2} + 2{y^2} – 10x – 12y + 12 = 0$

or ${S_2} = {x^2} + {y^2} – 5x – 6y + 6 = 0$ …..$(ii)$

$\therefore $ Equation of radical axis is ${S_1} – {S_2} = 0$

$ \Rightarrow $ $2x + 2y – 1 = 0$.

Standard 11

Mathematics

Share

Similar Questions

Figure shows $\Delta ABC$ with $AB = 3, AC = 4$ & $BC = 5$. Three circles $S_1, S_2$ & $S_3$ have their centres on $A, B $ & $C$ respectively and they externally touches each other. The sum of areas of three circles is

normal

A variable line $ax + by + c = 0$, where $a, b, c$ are in $A.P.$, is normal to a circle $(x – \alpha)^2 + (y – \beta)^2 = \gamma$ , which is orthogonal to circle $x^2 + y^2- 4x- 4y-1 = 0$. The value of $\alpha + \beta + \gamma$ is equal to

normal

The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0) , (1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are :

normal

Three circles of radii $a, b, c\, ( a < b < c )$ touch each other externally. If they have $x -$ axis as a common tangent, then

hard

(JEE MAIN-2019)

The equation of a circle that intersects the circle ${x^2} + {y^2} + 14x + 6y + 2 = 0$orthogonally and whose centre is $(0, 2)$ is

hard