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Question: The abscissa of two points A and B are the roots of the equation \({{x}^{2}}+2ax-{{b}^{2}}=0\) and t...

The abscissa of two points A and B are the roots of the equation x2+2ax−b2=0{{x}^{2}}+2ax-{{b}^{2}}=0 and their ordinates are the roots of the equation x2+2px−q2=0{{x}^{2}}+2px-{{q}^{2}}=0. The radius of the circle with AB as a diameter will be
(a) a2+b2+p2+q2\sqrt{{{a}^{2}}+{{b}^{2}}+{{p}^{2}}+{{q}^{2}}}
(b) b2+q2\sqrt{{{b}^{2}}+{{q}^{2}}}
(c) a2+b2−p2−q2\sqrt{{{a}^{2}}+{{b}^{2}}-{{p}^{2}}-{{q}^{2}}}
(d) a2+p2\sqrt{{{a}^{2}}+{{p}^{2}}}

Explanation

Solution

Hint: Find the coordinates of points A and B by using the sum of roots and product of roots formula of a quadratic equation. Then apply a distance formula to find the radius of the circle, i.e.,AB2\dfrac{AB}{2}.

Complete step-by-step answer:
First we have two find the roots of the equations.
Let the coordinate of A be (α,β).\left( \alpha ,\beta \right).
α,β\alpha ,\beta are the roots of the equation x2+2ax−b2.{{x}^{2}}+2ax-{{b}^{2}}.
We know, the product of the roots of a quadratic equation are given as,
α.β=Constant termCoefficient of x2\alpha .\beta =\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}
αβ=−b21=−b2......(i)\alpha \beta =\dfrac{-{{b}^{2}}}{1}=-{{b}^{2}}......(i)
The sum of the roots is given by,
(α+β)=−(Coefficient of x)Coefficient of x2\left( \alpha +\beta \right)=\dfrac{-(\text{Coefficient of }x)}{\text{Coefficient of }{{x}^{2}}}
(α+β)=−2a1=−2a........(ii)\left( \alpha +\beta \right)=\dfrac{-2a}{1}=-2a........(ii)
Again,
Let the coordinate of B is (α1,β1).\left( {{\alpha }_{1}},{{\beta }_{1}} \right).
α1,β1{{\alpha }_{1}},{{\beta }_{1}}are the roots of the equation x2+2px−q2=0.{{x}^{2}}+2px-{{q}^{2}}=0.
So, the product of roots in this case is,
α1β1=Constant termCoefficient of x2{{\alpha }_{1}}{{\beta }_{1}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{x}^{2}}}
α1β1=−q21=−q2.......(iii){{\alpha }_{1}}{{\beta }_{1}}=\dfrac{-{{q}^{2}}}{1}=-{{q}^{2}}.......(iii)
And the sum of the roots will be,
(α1+β1)=−(Coefficient of x)Coefficient of x2({{\alpha }_{1}}+{{\beta }_{1}})=\dfrac{-(\text{Coefficient of }x)}{\text{Coefficient of }{{x}^{2}}}
(α1+β1)=−2p1=−2p.........(iv)({{\alpha }_{1}}+{{\beta }_{1}})=\dfrac{-2p}{1}=-2p.........(iv)
To find the length AB (which is the diameter), we are applying the distance formula, i.e.,
(x−x1)2+(y−y1)2\sqrt{{{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}}
In this case (x,y)=(α,α1),(x1,y1)=(β,β1)(x,y)=(\alpha ,{{\alpha }_{1}}),({{x}_{1}},{{y}_{1}})=(\beta ,{{\beta }_{1}}), substituting the values, we get
AB=(α−β)2+(α1−β1)2AB=\sqrt{{{(\alpha -\beta )}^{2}}+{{({{\alpha }_{1}}-{{\beta }_{1}})}^{2}}}
Now we will try to convert this into addition as we know the respective values, we know the formula, (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} , using this in above expression, we get
AB=α2+β2−2αβ+α12+β12−2α1β1AB=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +\alpha _{1}^{2}+\beta _{1}^{2}-2{{\alpha }_{1}}{{\beta }_{1}}}
Adding and subtracting by (2αβ,2α1β1)(2\alpha \beta ,2{{\alpha }_{1}}{{\beta }_{1}}) , we get
AB=α2+β2+2αβ−2αβ−2αβ+α12+β12+2α1β1−2α1β1−2α1β1AB=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2\alpha \beta -2\alpha \beta +\alpha _{1}^{2}+\beta _{1}^{2}+2{{\alpha }_{1}}{{\beta }_{1}}-2{{\alpha }_{1}}{{\beta }_{1}}-2{{\alpha }_{1}}{{\beta }_{1}}}
Now regrouping, we get
AB=(α2+β2+2αβ)−4αβ+(α12+β12+2α1β1)−4α1β1AB=\sqrt{\left( {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \right)-4\alpha \beta +\left( \alpha _{1}^{2}+\beta _{1}^{2}+2{{\alpha }_{1}}{{\beta }_{1}} \right)-4{{\alpha }_{1}}{{\beta }_{1}}}
Now we know the formula, (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} , using this in above expression, we get
AB=(α+β)2−4αβ+(α1+β1)2−4α1β1AB=\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta +{{({{\alpha }_{1}}+{{\beta }_{1}})}^{2}}-4{{\alpha }_{1}}{{\beta }_{1}}}
Substituting the values from equation (i), (ii), (iii) and (iv), we get
AB=(−2a)2−4×(−b2)+(−2p)2−4×(−q2)AB=\sqrt{{{(-2a)}^{2}}-4\times (-{{b}^{2}})+{{(-2p)}^{2}}-4\times (-{{q}^{2}})}
AB=4a2+4b2+4p2+4q2AB=\sqrt{4{{a}^{2}}+4{{b}^{2}}+4{{p}^{2}}+4{{q}^{2}}}
Taking the common term out, we get
AB=4(a2+b2+p2+q2)AB=\sqrt{4({{a}^{2}}+{{b}^{2}}+{{p}^{2}}+{{q}^{2}})}
AB=2a2+b2+p2+q2AB=2\sqrt{{{a}^{2}}+{{b}^{2}}+{{p}^{2}}+{{q}^{2}}}
We have been told to find the radius, and AB is the diameter. We know radius is AB2.\dfrac{AB}{2}.
So, above equation becomes,
Radius=2a2+b2+p2+q22Radius=\dfrac{2\sqrt{{{a}^{2}}+{{b}^{2}}+{{p}^{2}}+{{q}^{2}}}}{2}
Radius=a2+b2+p2+q2Radius=\sqrt{{{a}^{2}}+{{b}^{2}}+{{p}^{2}}+{{q}^{2}}}
Hence the correct answer is option (a).

Note: Sum and product of roots needs to be calculated. Applying distance formula because it has been asked to find the radius of AB (whenever any distance or length is asked to find out).
The possible mistake that can be made here is in a hurry the students might directly substitute the value in AB=(α−β)2+(α1−β1)2AB=\sqrt{{{(\alpha -\beta )}^{2}}+{{({{\alpha }_{1}}-{{\beta }_{1}})}^{2}}}, instead of converting it to sum. And also the student forgot to divide the value of AB by two to get radius.