Radical centre of circles {x^2} + {y^2} - 16x + 60 = 0,,{x^2} + {y^2} - 12x + 27 = 0, {x^2} + {y^2}

English

Gujarati

Hindi

10-1.Circle and System of Circles

hard

The radical centre of the circles ${x^2} + {y^2} - 16x + 60 = 0,\,{x^2} + {y^2} - 12x + 27 = 0,$ ${x^2} + {y^2} - 12y + 8 = 0$ is

A

$(13, 33/4)$

B

$(33/4, -13)$

C

$(33/4, 13)$

D

None of these

Solution

(d) ${S_1}$$ \equiv $ ${x^2} + {y^2} – 16x + 60 = 0$ …..$(i)$

${S_2}$$ \equiv $${x^2} + {y^2} – 12x + 27 = 0$ …..$(ii)$

${S_3}$$ \equiv $${x^2} + {y^2} – 12y + 8 = 0$ …..$(iii)$

The radical axis of circle $(i)$ and circle $(ii)$ is

${S_1} – {S_2} = 0\, \Rightarrow \, – 4x + 33 = 0$ ….$(iv)$

the radical axis of circle $(ii)$ and circle $(iii)$ is ${S_2} – {S_3} = 0$

$ \Rightarrow \,\, – 12 + 12y + 19 = 0$ …..$(v)$

Solving $(iv)$ and $(v),$ we get the radical centre $\left( {\frac{{33}}{4},\,\frac{{20}}{3}} \right)$.

Standard 11

Mathematics

Share

Similar Questions

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

medium

The equation of the circle which passes through the origin, has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} – 4x + 2y + 4 = 0$ orthogonally, is

hard

The radical axis of the pair of circle ${x^2} + {y^2} = 144$ and ${x^2} + {y^2} – 15x + 12y = 0$ is

medium

Answer the following by appropriately matching the lists based on the information given in the paragraph

Let the circles $C_1: x^2+y^2=9$ and $C_2:(x-3)^2+(y-4)^2=16$, intersect at the points $X$ and $Y$. Suppose that another circle $C_3:(x-h)^2+(y-k)^2=r^2$ satisfies the following conditions :

$(i)$ centre of $C _3$ is collinear with the centres of $C _1$ and $C _2$

$(ii)$ $C _1$ and $C _2$ both lie inside $C _3$, and

$(iii)$ $C _3$ touches $C _1$ at $M$ and $C _2$ at $N$.

Let the line through $X$ and $Y$ intersect $C _3$ at $Z$ and $W$, and let a common tangent of $C _1$ and $C _3$ be a tangent to the parabola $x^2=8 \alpha y$.

There are some expression given in the $List-I$ whose values are given in $List-II$ below:

$List-I$ $List-II$

$(I)$ $2 h + k$ $(P)$ $6$

$(II)$ $\frac{\text { Length of } ZW }{\text { Length of } XY }$ $(Q)$ $\sqrt{6}$

$(III)$ $\frac{\text { Area of triangle } MZN }{\text { Area of triangle ZMW }}$ $(R)$ $\frac{5}{4}$

$(IV)$ $\alpha$ $(S)$ $\frac{21}{5}$

$(T)$ $2 \sqrt{6}$

$(U)$ $\frac{10}{3}$

($1$) Which of the following is the only INCORRECT combination?

$(1) (IV), (S)$ $(2) (IV), (U)$ $(3) (III), (R)$ $(4) (I), (P)$

($2$) Which of the following is the only CORRECT combination?

$(1) (II), (T)$ $(2) (I), (S)$ $(3) (I), (U)$ $(4) (II), (Q)$

Give the answer or quetion ($1$) and ($2$)

hard

(IIT-2019)

The locus of centre of the circle which cuts the circles${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ orthogonally is

normal