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Question

The centre of a circle is $$(2a, a-7$$). Find the values of $$a$$ if the circle passes through the point (11, -9) and has diameter $$10\sqrt{2}$$ units.


A
35
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B
3
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C
5
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D
53
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Solution

The correct options are
A $$\dfrac 35$$
C $$ 5 $$
$$ Let\quad the\quad given\quad point\quad P\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 11,-9 \right) \quad lie\quad on\quad a\quad circle\quad with\quad centre\quad at\quad $$

$$ C\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 2a,a-7 \right) \quad \& \quad radius=r.$$

$$ Then\quad PC=r.......(i)\quad $$

$$ Given\quad that\quad the\quad diameter\quad of\quad the\quad circle=10\sqrt { 2 } units.$$

$$ \therefore \quad r=\dfrac { diameter }{ 2 } =\dfrac { 10\sqrt { 2 } units }{ 2 } =5\sqrt { 2 } units....(ii).$$

$$ Again,\quad by\quad distance\quad formula,\quad d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right)  }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right)  }^{ 2 } } \quad $$

$$ \therefore \quad PC=d=\sqrt { { \left( 11-2a \right)  }^{ 2 }+{ \left( -9-a+7 \right)  }^{ 2 } } =\sqrt { 5{ a }^{ 2 }-40a+125 } .$$

$$ \therefore \quad From\quad (i)\quad \& \quad (ii)\quad we\quad get $$

$$ 5{ a }^{ 2 }-40a+125=({ 5\sqrt { 2 } ) }^{ 2 }$$

$$ \Longrightarrow 5{ a }^{ 2 }-40a+75=0$$

$$ \Longrightarrow \left( a-5 \right) \left( 5a-3 \right) =0$$

$$ \Longrightarrow a=5,\dfrac { 3 }{ 5 } $$

$$ Ans\quad a=5,\dfrac { 3 }{ 5 }   $$


Mathematics

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