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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 :
$\left( {\frac{3}{2}\,\,,\,\,\frac{1}{2}} \right)$
$\left( {\frac{1}{2}\,\,,\,\, - \,{2^{1/2}}} \right)$
$\left( {\frac{1}{2}\,\,,\,\,{2^{1/2}}} \right)$
$(B)$ or $(C)$ both
Solution
consider family of $ \odot$ 's through $(0, 0)$ and $(1, 0)$
$x(x – 1) + y^2 + \lambda y = 0$
touches $x^2 + y^2 = 9$
$\therefore$ common chord $= – x + hy + 9 = 0….(1)$
$\therefore$ perpendicular from $(0, 0)$ on $(1)$ is equal to $3$.
$\left| {\,\frac{9}{{\sqrt {1 + {\lambda ^2}} }}\,} \right|$ $= 3$ $\Rightarrow \,\, \lambda_2 = 8$ $\Rightarrow \,\, \lambda = \pm 2\sqrt{2}$
circle $x (x – 1) + y^2 + 2\sqrt{2} y =0$
$\therefore$ centre $\left( {\frac{1}{2}\,\,,\,\, – \,{2^{1/2}}} \right)$
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 ($2$)
