The equation of the circle passing through the points 4-1-6-5 whose centre lies on the 4 x- y -16-0 isA- x2-y2-10 x-6 y-29-0B- x2-y2-x-13-0C- x2-y2-6 x-8 y-15-0D- x2-y2-4 y-12-0

The correct option is C x 2+y2 6x 8y+15=0 A(4,1) 4x+y 16 = 0 B(6,5) C(5,3) Slope of AB = (y/2) = 2 y 3 = 1/2(x 5) r = AC = √(1+9) = √(10) Slope = ( 1/ ...

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Standard XII

Mathematics

Length of Chord

The equation ...

Question

The equation of the circle passing through the points (4,1),(6,5) whose centre lies on the 4x+y-16=0 is

x²+y²-10x-6y+29=0

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x²+y²-x-13=0

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x²+y²-6x-8y+15=0

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x²+y²+4y-12=0

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Solution

The correct option is C
x²+y²-6x-8y+15=0

$A (4, 1)$ $4 x + y - 16$ = 0

$B (6, 5)$ $C (5, 3)$

Slope of AB = $(\frac{y}{2})$ = 2 $y - 3$ = $- \frac{1}{2} (x - 5)$ $r$ = AC = $\sqrt{1 + 9}$ = $\sqrt{10}$

$⊥$ Slope = ( $- \frac{1}{2}$ ) $2 y - 6$ = $- x + 5$ $x + 2 y - 11$ = 0 $(x - 3)^{2} + (y - 4)^{2}$ = 10

$\Rightarrow x^{2} + y^{2} - 6 x - 8 y + 15$ = 0

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The equation of the circle passing through the points (4,1),(6,5) whose centre lies on the 4x+y-16=0 is

The correct option is C x 2+y2-6x-8y+15=0 A(4,1) 4x+y−16 = 0 B(6,5) C(5,3) Slope of AB = (y2) = 2 y−3 = −12(x−5) r = AC = √1+9 = √10 ⊥ Slope = (−12) 2y−6 = −x+5 x+2y−11 = 0 (x−3)2+(y−4)2 = 10 ⇒x2+y2−6x−8y+15 = 0