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Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals
$110$
$130$
$125$
$145$
Solution
$ \because \frac{16+\alpha}{3}=6 \text { and } \frac{15+\beta}{3}=6 $
$ \Rightarrow(\alpha, \beta) \equiv(2,3) $
$ \text { Also, } \mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1+\mathrm{r}_2 $
$ \Rightarrow \sqrt{(2-8)^2+\left(3-\frac{15}{2}\right)^2}=2 \mathrm{r}_2+\mathrm{r}_2 $
$ \Rightarrow \mathrm{r}_2=\frac{5}{2} \Rightarrow \mathrm{r}_1=2 \mathrm{r}_2=5 $
$ \therefore(\alpha+\beta)+4\left(\mathrm{r}_1^2+\mathrm{r}_2^2\right) $
$ =5+4\left(\frac{25}{4}+25\right)=130$
Similar Questions
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 ($2$)
