Question

03 The center
of a circle is (2P-1,P). find
the value of p if circle is passes though
points (10,-2) & dia is 10√2​

Answer 1

Answer:

Here diameter

2r=10

2

r=5

2

OA=r=5

2

=

(2a−1−10)+(a+2)

2

=5

2

(2a−11)

2

+(a+2)

2

=50

4a

2

−44a+121+a

2

+4a+4=50

5a

2

−40a+75=0

a

2

−8a+15=0

a−5a−3a+15=0

(a−5a)(a−3)=0

a=5or3

Answer 2

Given, diameter of the circle is 10√2 units.

Hence the distance between OB = Radius of the circle.

We have to find, the value of p.

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$\setlength{\unitlength}{1 mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\circle*{1}}\put(24,32){\sf\large{O}}\put(21,24){\sf\large{(2p-1,p)}}\put(45,30){\line(-1,0){40}}\put(25,30){\circle*{1}}\put(46,30){\sf\large{B}}\put(46,24){\sf\large{}}\put(45,30){\circle*{1}}\put(1,31){\sf\large{A}}\put(5,30){\circle*{1}}\put(-6,24){\sf\large{}}\end{picture}$

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★ Coordinate of centre of circle = (2p-1,p).

$:\implies\sf \sqrt { 10 - (2p - 1)^{2} + { (- 2 - p)}^{2} } = 5 \sqrt{2}$

By squaring on both the sides,

$:\implies\sf {(11 - 2p)}^{2} + {(4 + {p}^{2} + 4p)} = 25 \times 2$

$:\implies\sf 121 + 4p ^{2} - 44p + 4 + p^{2} = 50$

$:\implies\sf {5p}^{2} - 40p = \: 50 - 125$

$:\implies\sf {5p}^{2} - 40p + 75 = 0$

$:\implies\sf 5( {p}^{2} - 8p + 15) = 0$

$:\implies\sf {p}^{2} - (15 + 3)p + 15 = 0$

$:\implies\sf p(p - 5) - 3(p - 5) = 0$

$:\implies\sf ( p- 5)( p- 3) = 0$

$:\implies\sf p-5=0$

$:\implies\sf p-3=0$

$:\implies{\boxed{\frak{\pink{p=5\;or\;3}}}}\;\bigstar$

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