Show that the equation of any circle passing through the points of intersection of the ellipse (x + 2)² + 2y² = 18 and the ellipse 9(x - 1)² + 16y² = 25 can be written in the form x² - 2ax + y² = 5 - 4a.
Answers
Given : A circle passes through the points of intersection of the ellipse (x + 2)² + 2y² = 18 and the ellipse 9(x - 1)² + 16y² = 25.
To prove : The equation of the circle can be written in the form x² - 2ax + y² = 5 - 4a.
Solution :
Firstly we have to find the intersecting points of both ellipses which can be done easily by substituting value of one variable into other.
Solving the equation (x+2)² + 2y² = 18 for y² :-
⇒ (x+2)² + 2y² = 18
⇒ 2y² = 18 - (x+2)²
⇒ y² = [18 - (x+2)²]/2
Now substitute this value of y in second eq. :-
⇒ 9(x-1)² + 16y² = 25
⇒ 9(x-1)² + 16 [18 - (x+2)²]/2 = 25
⇒ 9(x-1)² + 8 [18 - (x+2)²] = 25
⇒ 9[x² + 1 - 2x] + 8 [18 - (x² + 4x + 4)] = 25
⇒ 9[x² + 1 - 2x] + 8 [18 - x² - 4x - 4] = 25
⇒ 9x² + 9 - 18x + 144 - 8x² - 32x - 32 = 25
⇒ x² - 50x + 96 = 0
By solving this quad. equation, we will get,
- x = 2 and x = 48
This x = 48 will be rejected ( Hint : see the attachment to know why would x = 48 will be rejected ).
Therefore x = 2
Now substitute x = 2 in first eq. :-
⇒ (x + 2)² + 2y² = 18
⇒ (2 + 2)² + 2y² = 18
⇒ 4² + 2y² = 18
⇒ 16 + 2y² = 18
⇒ 2y² = 18 - 16
⇒ 2y² = 2
⇒ y = ±1
Therefore the intersection points are (2, 1) and (2, -1). So the circle will pàss through these points.
Now, we know that the center of the circle will lie on x axis.
Let's assume that the center is (a, 0).
So the eq. of the circle is given by,
⇒ (x - a)² + y² = r²
⇒ x² - 2ax + a² + y² = r² ______(1)
Now we know that the circle passes through (2, 1) and (2, -1). So by putting these values for x and y will satisfy the equation.
Therefore,
⇒ x² - 2ax + a² + y² = r²
⇒ (2)² - 2a(2) + a² + (-1)² = r²
⇒ 4 - 4a + a² + 1 = r²
⇒ a² - 4a + 5 = r²
Substitute this value of r² in eq. (1)
⇒ x² - 2ax + a² + y² = r²
⇒ x² - 2ax + a² + y² = a² - 4a + 5
⇒ x² - 2ax + y² = 5 - 4a.
∴ The required result is proved.
Thanks for posting this question, I learnt the concept of circle and ellipse after you posted this question! :)
Answer:
First we have to find the intersections of the two ellipses by solving the simultaneous equations
(x+2)2+2y29(x−1)2+16y2==1825.(1)(2)
We can eliminate y by multiplying equation (1) by 8 and subtracting equation (2), so that at the intersections
8(x+2)2−9(x−1)2i.e.x2−50x+96i.e.(x−2)(x−48)=144−25=0=0.
The two possible values for x at the intersections are therefore 2 and 48.
Next we find the values of y at the intersection. Taking x=2 and substituting into equation (1) gives 16+2y2=18, so y=±1. Taking x=48 gives 502+2y2=18 which has no (real) roots. Thus there are two points of intersection, at (2,1) and (2,−1).
Now we go after the circle. Suppose that a circle through the points of intersection has centre (p,q) and radius R. Then the equation of the circle is
(x−p)2+(y−q)2=R2.
Setting (x,y)=(2,1) and (x,y)=(2,−1) gives two equations:.
(2−p)2+(1−q)2=R2,(2−p)2+(−1−q)2=R2.
Subtracting the two equations gives q=0 (it is obvious anyway, because of the symmetry of both ellipses under reflections in the x axis, that the centre of the circle must lie on the y axis). Thus the equation of any circle passing through the intersections is
(x−p)2+y2=(2−p)2+1,
which simplifies to the given result with p=a.