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Question

If (h,k) is the centre of the circle touches yaxis at a distance of 12 units from the origin and makes an intercept of 10 units on xaxis, then the equation of circle for which (h+k) is minimum, is

A
(x13)2+(y+12)2=169
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B
(x+13)2+(y12)2=169
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C
(x13)2+(y12)2=169
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D
(x+13)2+(y+12)2=169
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Solution

The correct option is D (x+13)2+(y+12)2=169
The centre of the circle is
C(h,±12)

Assuming the point of intersection of the circle with x axis at A and B, making a rough diagram, we get


In ABC
AC2=AM2+CM2r2=122+52=169r=13
Therefore the possible center of the circle are (±13,±12)
So, h+k is minimum when h=13,k=12

The required equation of the circle is
(x+13)2+(y+12)2=169

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