Choose the correct statement about two circles whose equations are given below :
x2 + y2 $-$ 10x $-$ 10y + 41 = 0
x2 + y2 $-$ 22x $-$ 10y + 137 = 0
Solution
Let ${S_1}:{x^2} + {y^2} - 10x - 10y + 41 = 0$<br><br>$\Rightarrow {(x - 5)^2} + {(y - 5)^2} = 9$<br><br>Centre $({C_1}) = (5,5)$<br><br>Radius r<sub>1</sub> = 3<br><br>${S_2}:{x^2} + {y^2} - 22x - 10y + 137 = 0$<br><br>$\Rightarrow {(x - 11)^2} + {(y - 5)^2} = 9$<br><br>Centre $({C_2}) = (11,5)$<br><br>Radius r<sub>2</sub> = 3<br><br>distance $({C_1}{C_2}) = \sqrt {{{(5 - 11)}^2} + {{(5 - 5)}^2}}$<br><br>distance $({C_1}{C_2}) = 6$<br><br>$\because$ ${r_1} + {r_2} = 3 + 3 = 6$<br><br>$\therefore$ circles touch externally<br><br>Hence, circle have only one meeting point.
About this question
Subject: Mathematics · Chapter: Circles · Topic: Radical Axis
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