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If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then
$2 < r < 8$
$r = 2$
$r < 2$
$r > 2$
Solution
(a) When two circles intersect each other, then
Difference between their radii $r – 3 < 5$ $⇒$ $r = 8$…..$(i)$
Difference between their radiiSum of their radii > Distance between centres ….$(ii)$
$ \Rightarrow $ $r + 3 > 5$
$\Rightarrow r > 2$
Hence by (i) and (ii), $2 < r < 8$.
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 ($2$)