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Question: If (a, c) and (b, c) are the centres of the two circles whose radial axis is y-axis. If the radius o...

If (a, c) and (b, c) are the centres of the two circles whose radial axis is y-axis. If the radius of the first circle is r then the diameter of the other circle is
(A)2r2−b2+a2 (B)r2−a2+b2 (C)(r2−b2+a2) (D)2r2−a2+b2   \left( {\text{A}} \right)2\sqrt {{r^2} - {b^2} + {a^2}} \\\ \left( {\text{B}} \right)\sqrt {{r^2} - {a^2} + {b^2}} \\\ \left( {\text{C}} \right)\left( {{r^2} - {b^2} + {a^2}} \right) \\\ \left( {\text{D}} \right)2\sqrt {{r^2} - {a^2} + {b^2}} \;

Explanation

Solution

Hint : Start the solution by naming both the circles and their respective radius. Write the general equation of the circles individually. After this use the formula of radial axis and make use of the information given about the radial axis in the question. From this you will obtain the radius and hence the diameter.

Complete step-by-step answer :
We have been given two circles with centres (a, c) and (b, c).
Let us name these circles as A and B respectively.
Also let circle A have radius r as told and let circle B have radius R respectively.
Now the general equation of the circle A will be
(x−a)2+(y−c)2=r2{\left( {x - a} \right)^2} + {\left( {y - c} \right)^2} = {r^2}
And the general equation of circle B will be
(x−b)2+(y−c)2=R2{\left( {x - b} \right)^2} + {\left( {y - c} \right)^2} = {R^2}
The radial axis given to us is y- axis i.e. x=0.
Hence by the formula of radial axis,
A - B = x = [(x−a)2+(y−c)2−(x−b)2−(y−c)2−r2+R2]{\text{A - B = x = }}\left[ {{{\left( {x - a} \right)}^2} + {{\left( {y - c} \right)}^2} - {{\left( {x - b} \right)}^2} - {{\left( {y - c} \right)}^2} - {r^2} + {R^2}} \right]
=[(x−a)2−(x−b)2−r2+R2] =[x2−2ax+a2−(x2−2bx+b2)−r2+R2] =[a2−b2−2ax+2bx−r2+R2]   = \left[ {{{\left( {x - a} \right)}^2} - {{\left( {x - b} \right)}^2} - {r^2} + {R^2}} \right] \\\ = \left[ {{x^2} - 2ax + {a^2} - \left( {{x^2} - 2bx + {b^2}} \right) - {r^2} + {R^2}} \right] \\\ = \left[ {{a^2} - {b^2} - 2ax + 2bx - {r^2} + {R^2}} \right] \;
Hence x=[a2−b2−2ax+2bx−r2+R2]{\text{x}} = \left[ {{a^2} - {b^2} - 2ax + 2bx - {r^2} + {R^2}} \right]
Substituting x=0 we get
0 = [R2−r2+a2−b2] ⇒R2=r2+b2−a2 ⇒R=r2+b2−a2   {\text{0 = }}\left[ {{R^2} - {r^2} + {a^2} - {b^2}} \right] \\\ \Rightarrow {R^2} = {r^2} + {b^2} - {a^2} \\\ \Rightarrow R = \sqrt {{r^2} + {b^2} - {a^2}} \;
Since this is the radius, diameter = 2. Radius
Hence, D=2r2+b2−a2D = 2\sqrt {{r^2} + {b^2} - {a^2}}
Hence the right option to the question is option D.
So, the correct answer is “Option D”.

Note : The radial axis is a line perpendicular to the line joining the centres of the two circles. Also the radical axis of three circles taken in pairs meeting at an appointment is called their radical centre. The coordinate of the radical circle can be found by solving the equation S1=S2=S3=0{S_1} = {S_2} = {S_3} = 0 and the radical centre of three circle having sides of triangles as diameter is the orthocentre of the triangle.