Question

The coordinates of the point A, where AB is the diameter of the circle with center O(1,3) and B(5,4) are: (A) (-3,2) (B) Q(4,-9)(C) Q(2,-3) (D) Q(3,-4)

Answer 1

        $\text{\huge \bf \underline{Answer:}}$

        $\text{It is given that }AB\text{ is the diameter of a circle, where }B(5,4)\\O \text{ is the center of the circle with coordinates }(1,3)$

        $\text{Let }A \text{ be }(x,y)$

        $\text{We know that twice the radius is the length of diameter}$

$\implies \: 2OA = AB \text{ ------ (i)}$

        $\text{Applying distance formula: }\sqrt{(x_1-x_2)^{2} + (y_1 - y_2)^{2}}$

$\implies \: (x_1,y_1) = A(x,y) \: \: ; \: \: (x_2,y_2) = O(1,3)$

$\implies \: OA = \sqrt{(x-1)^{2} + (y-3)^{2} }$

        $\text{We know that }(a-b)^{2} = a^{2} + b^{2} - 2ab$

$\implies \: OA = \sqrt{(x)^{2} + (1)^{2} - 2(1)x + (y)^{2} + (3)^{2} - 2(3)y}$

$\implies \: OA = \sqrt{x^{2} + 1 - 2x + y^{2} + 9 - 6y}$

$\implies \: OA = \sqrt{x^{2} + y^{2} - 2x - 6y + 10 }$

        $\text{Similarly, }AB = \sqrt{(x - 5)^{2} + (y-4)^{2} }$

        $\text{We know that }(a-b)^{2} = a^{2} + b^{2} - 2ab$

$\implies \: AB = \sqrt{(x)^{2} + (5)^{2} - 2(5)x + (y)^{2} + (4)^{2} - 2(4)y }$

$\implies \: AB = \sqrt{x^{2} + 25 - 10x + y^{2} + 16 - 8y }$

$\implies \: AB = \sqrt{x^{2} + y^{2} - 10x - 8y + 41}$

        $\text{Substituting in (i)}$

$\implies \: 2(\sqrt{x^{2} + y^{2} - 2x - 6y + 10}) = \sqrt{x^{2} + y^{2} - 10x - 8y + 41}$

        $\text{Squaring both sides}$

$\implies \: (2)^{2} (x^{2} + y^{2} - 2x - 6y + 10) = x^{2} + y^{2} - 10x - 8y + 41$

$\implies \: 4 (x^{2} + y^{2} - 2x - 6y + 10) = x^{2} + y^{2} - 10x - 8y + 41$

$\implies \: 4x^{2} + 4y^{2} - 8x - 24y + 40 = x^{2} + y^{2} - 10x - 8y + 41$

$\implies \: 4x^{2} + 4y^{2} - 8x - 24y + 40 - x^{2} - y^{2} + 10x + 8y - 41 = 0$

$\implies \: 3x^{2} + 3y^{2} +2x - 16y - 1 = 0$

        $\text{We have obtained an equation of point }A$

        $\text{Since we have been given options we're going to substitute each}\\\text{one into the equation and check whether it's the solution or not}\\\text{if it is, then the substituted value is the correct option.}$

        $A) (-3,2)$

        $\text{Substituting in the equation we get}$

$\implies \: 3(-3)^{2} + 3(2)^{2} +2(-3) - 16(2) - 1 = 0$

$\implies \: 3(9) + 3(4) -6 - 32 - 1 = 0$

$\implies \: 27 + 12 -6 - 32 - 1 = 0$

$\implies \: 39 -39 = 0$

$\implies \: 0 = 0$

$\implies \: \text{L.H.S = R.H.S}$

$\implies \: \boxed{\boxed{\text{The coordinates of }A = (-3,2)}}$