Question
Question: The equation of the circle passing through the origin and through the points of intersection of circ...
The equation of the circle passing through the origin and through the points of intersection of circles x2+y2=4 and x2+y2−2x−4y+4=0
Solution
In this particular question use the concept that the equation of circle passing through the intersection of two circles is gives as, x2+y2−4+K(x2+y2−2x−4y+4)=0, where K is any real parameter, so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given equation of circles
x2+y2=4........................ (1)
And
x2+y2−2x−4y+4=0..................... (2)
Now we have to find the equation of the circle passing through the point of intersection.
So the equation of circle is given as,
⇒x2+y2−4+K(x2+y2−2x−4y+4)=0....................... (1), where K is any real parameter.
Now it is also given that it is passing through origin (0, 0) so it satisfies the above equation, so substitute this value i.e. x = y = 0, in the above equation, so we have,
⇒02+02−4+K(02+02−2(0)−4(0)+4)=0
Now simplify this equation we have,
⇒−4+K(4)=0
⇒K(4)=4
⇒K=1
Now substitute this value in equation (1) we have,
⇒x2+y2−4+1(x2+y2−2x−4y+4)=0
Now simplify the above equation we have,
⇒2x2+2y2−4−2x−4y+4=0
⇒2x2+2y2−2x−4y=0
Now divide by 2 throughout we have,
⇒x2+y2−x−2y=0
So this is the required equation of the circle.
So this is the required answer.
Note: Whenever we face such types of questions the key concept we have to remember is that if a circle is passing through any point say (h, k) then these points i.e. x = h, y = k satisfies the equation of circle, so first find out the equation of circle passing through the intersection of two circles as above then satisfies the coordinates of origin in this equation we will get the value of parameter K as above.