Let C₁ be circle of radius 1 with center at origin. Let C₂ be circle of radius mathrm{I} with ce

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10-1.Circle and System of Circles

medium

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $\mathrm{I}$ with center at the point $A=(4,1)$, where $1<\mathrm{r}<3$. Two distinct common tangents $P Q$ and $S T$ of $C_1$ and $C_2$ are drawn. The tangent $P Q$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $S T$ touches $C_1$ at $S$ and $C_2$ at $T$. Mid points of the line segments $P Q$ and $S T$ are joined to form a line which meets the $x$-axis at a point $B$. If $A B=\sqrt{5}$, then the value of $r^2$ is

$2$

$5$

$8$

$7$

(IIT-2023)

Solution

Let $C_2(x-4)^2+(y-1)^2=r^2$ radical axis $8 x+2 y-17=1-r^2$ $8 \mathrm{x}+2 \mathrm{y}=18-\mathrm{r}^2$ $\mathrm{B}\left(\frac{18-\mathrm{r}^2}{8}, 0\right) \mathrm{A}(4,1)$ $\mathrm{AB}=\sqrt{5}$ $\sqrt{\left(\frac{18-r^2}{8}-4\right)^2+1}=\sqrt{5}$ $r^2=2$ $\Rightarrow \mathrm{n}=\sin \alpha+\cos \alpha$

Standard 11

Mathematics

If the circle ${x^2} + {y^2} + 6x – 2y + k = 0$ bisects the circumference of the circle ${x^2} + {y^2} + 2x – 6y – 15 = 0,$ then $k =$

hard

The two circles ${x^2} + {y^2} – 2x – 3 = 0$ and ${x^2} + {y^2} – 4x – 6y – 8 = 0$ are such that

medium

In the figure shown, radius of circle $C_1$ be $ r$ and that of $C_2$ be $\frac{r}{2}$ , where $r= \frac {1}{3} PQ,$ then length of $AB$ is (where $P$ and $Q$ being centres of $C_1$ $\&$ $C_2$ respectively)

normal

The equation of the image of the circle ${x^2} + {y^2} + 16x – 24y + 183 = 0$ by the line mirror $4x + 7y + 13 = 0$ is