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Question: The circle through (-2, 5), (0, 0) and intersecting the circle \[{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{...

The circle through (-2, 5), (0, 0) and intersecting the circle x2+y2−4x+3y−2=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ - 4x + 3y - 2 = 0}} orthogonally is
A) 2x2+2y2−11x−16y=02{{\rm{x}}^2}{\rm{ + 2}}{{\rm{y}}^2} - 11{\rm{x - 16y = 0}}
B) x2+y2−4x−4y=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2} - 4{\rm{x - 4y = 0}}
C) x2+y2+2x−5y=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ + 2x - 5y = 0}}
D) x2+y2+2x−5y+1=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ + 2x - 5y + 1 = 0}}

Explanation

Solution

Here, we have to use the concept of orthogonal of circles and its properties to find out the equation of the circle. Firstly we have to find the equation of the circle passing through the given points and then find another equation using the orthogonal property of the circle. Then by solving these two equations we will be able to find out the equation of the circle.

Complete step by step solution:
It is given that the circle passes through the points (-2, 5), (0, 0).
Let x2+y2+2gx+2fy=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ + 2gx + 2fy = 0}}………. (1) is the equation of the circle which passes through these points.
It is also given that the circle intersects this circle x2+y2−4x+3y−2=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ - 4x + 3y - 2 = 0}} orthogonally. So, firstly we will find the equation of the circle which passes through the given points, we get
(−g+2)2+(−f−5)2=g2+f2\sqrt {{{( - {\rm{g}} + 2)}^2} + {{( - {\rm{f}} - 5)}^2}} = \sqrt {{{\rm{g}}^2}{\rm{ + }}{{\rm{f}}^2}}
Simplifying the above equation, we get
⇒g2+f2−4g+4+10f+25=g2+f2\Rightarrow {{\rm{g}}^2}{\rm{ + }}{{\rm{f}}^2} - 4{\rm{g}} + 4 + 10{\rm{f}} + 25 = {{\rm{g}}^2}{\rm{ + }}{{\rm{f}}^2}
⇒10f−4g+29=0\Rightarrow 10{\rm{f}} - 4{\rm{g}} + 29 = 0 ………. (2)
Now we have to find another equation by using the orthogonal property of the circle i.e.
⇒2g1g2+2f1f2=c1+c2\Rightarrow 2{{\rm{g}}_1}{{\rm{g}}_2} + 2{{\rm{f}}_1}{{\rm{f}}_2} = {{\rm{c}}_1} + {{\rm{c}}_2}
In the above equation value of g2,f2andc2{{\rm{g}}_2}{\rm{,}}{{\rm{f}}_2}{\rm{ and }}{{\rm{c}}_2} is obtained by comparing the assumed equation (1) of the circle with the intersecting circle equation, we get
g2=−2,f2=32andc2=−1{{\rm{g}}_2}{\rm{ = - 2, }}{{\rm{f}}_2} = \dfrac{3}{2}{\rm{ and }}{{\rm{c}}_2} = - 1
Therefore equation becomes
⇒2g1(−2)+2f1(32)=−1\Rightarrow 2{{\rm{g}}_1}( - 2) + 2{{\rm{f}}_1}\left( {\dfrac{3}{2}} \right) = - 1
⇒−4g+3f=−1\Rightarrow - 4{\rm{g}} + 3{\rm{f}} = - 1 ………. (3)
Now, by solving equation (2) and (3) we get the value of g and f and by putting the value of g and f in the assumed equation (1) of the circle, we will get the equation of the circle.
Therefore solving equation (2) and (3), we get
f=−4andg=−114{\rm{f}} = - 4{\rm{ and g = }}\dfrac{{ - 11}}{4}
Now putting the value of g and f in equation (1), we get
x2+y2+2(−114)x+2(−4)y=0{{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ + 2}}\left( {\dfrac{{ - 11}}{4}} \right){\rm{x + 2( - 4)y = 0}}
By simplification we get
⇒2x2+2y2−11x−16y=0\Rightarrow 2{{\rm{x}}^2}{\rm{ + 2}}{{\rm{y}}^2} - 11{\rm{x - 16y = 0}}

So option A is correct.

Note:
Two circles are said to be orthogonal if the tangents to the both circles at the point of intersection makes 90∘{90^{\circ}} angle with each other i.e. perpendicular to each other then circles are aid that they are orthogonal to each other. Any symmetrical shape can intersect any other symmetric shape orthogonally. We should know the basic equation of the circle passing through a point.
Equation for the orthogonal property of the circle is 2g1g2+2f1f2=c1+c22{{\rm{g}}_1}{{\rm{g}}_2} + 2{{\rm{f}}_1}{{\rm{f}}_2} = {{\rm{c}}_1} + {{\rm{c}}_2}, where variables with suffix 1 is for 1st circle and variables with suffix 2 is for 2nd circle.