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NCERT Exemplar · Q53

Q.The shortest distance from the point (2,−7)(2, -7) to the circle x2+y2−14x−10y−151=0x^2 + y^2 - 14x - 10y - 151 = 0 is equal to 5.

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The center is (7,5)(7,5) and radius 1515; the point (2,−7)(2,-7) lies 1313 units from the center, so it is inside the circle and the shortest distance is 15−13=215-13=2, not 55. The statement is false.

For a point and a circle, the shortest distance is measured along the line through the center.

1. Center and radius. Complete the square on x2+y2−14x−10y−151=0x^2+y^2-14x-10y-151=0:

(x−7)2+(y−5)2=151+49+25=225(x-7)^2+(y-5)^2=151+49+25=225

So center C=(7,5)C=(7,5) and radius r=225=15r=\sqrt{225}=15.

2. Distance from the point to the center.

d=(7−2)2+(5+7)2=25+144=169=13d=\sqrt{(7-2)^2+(5+7)^2}=\sqrt{25+144}=\sqrt{169}=13 …

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