Centre of smallest circle touching circles x^2 + y^2- 2y - 3 = 0 and x^2+ y^2 - 8x

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10-1.Circle and System of Circles

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The centre of the smallest circle touching the circles $x^2 + y^2- 2y - 3 = 0$ and $x^2+ y^2 - 8x - 18y + 93 = 0$ is :

$(3 , 2)$

$(4 , 4)$

$(2 , 7)$

$(2 , 5)$

Solution

let

$S_{1}=x^{2}+y^{2}-2 y-3=0$

$C_{1}(0,1), r_{1}=2$

$S_{2}: x^{2}+y^{2}-8 x-18 y+93=0$

$C_{2}(4,9), r_{2}=2$

Hence, for smallest circle ${ }^{\prime} C_{3}^{\prime}$ centre is midpoint of ${ }^{\prime} C_{1}^{\prime},{ }^{\prime} C_{2}^{\prime}$ $\therefore C_{3}\left(\frac{0+4}{2}, \frac{9+1}{2}\right)$

$C_{3}(2,5)$

Standard 11

Mathematics

If ${x^2} + {y^2} + px + 3y – 5 = 0$ and ${x^2} + {y^2} + 5x$ $ + py + 7 = 0$ cut orthogonally, then $p$ is

medium

Let

$A=\left\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\right\}$

$B=\left\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\right\} \text { and }$

$C=\left\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\right\}$

Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to:

hard

(JEE MAIN-2021)

The two circles ${x^2} + {y^2} – 4y = 0$ and ${x^2} + {y^2} – 8y = 0$

medium

If a circle passes through the point $(1, 2)$ and cuts the circle ${x^2} + {y^2} = 4$ orthogonally, then the equation of the locus of its centre is

hard

The number of common tangents to the circles ${x^2} + {y^2} = 4$ and ${x^2} + {y^2} – 6x – 8y = 24$ is

hard

(IIT-1998)

The centre of the smallest circle touching the circles $x^2 + y^2- 2y - 3 = 0$ and $x^2+ y^2 - 8x - 18y + 93 = 0$ is :

Solution

Similar Questions

If ${x^2} + {y^2} + px + 3y – 5 = 0$ and ${x^2} + {y^2} + 5x$ $ + py + 7 = 0$ cut orthogonally, then $p$ is

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