Question

TYTULUY TOU.
Show that the circles x² + y2 –2x+4y+4= 0 and x' + y2 +3x+4y+1=0 are intersect
each other orthogonally.​

Answer 1

It's too simple dear

Orthogonal intersection means that the tangents at the point of the intersection are perpendicular to each other

However rather than going into a complex solution I think you should know the simple condition for two circles to intersect orthogonally

Let me derive it for you

Circle 1 ie C1 :-

x^2 + y^2 + 2g1x + 2f1y + c1 = 0

Circle 2 ie C2 :-

x^2 + y^2 + 2g2x + 2f2y + c2 = 0

Now if the tangents are perpendicular at the points of intersection of two Orthogonal circles then the normal at these points are also perpendicular

Now you already know that Normals to a circle always pass through center of the circle

So for the normal they pass through CENTER and then intersect the common chord of the two Orthogonal circles

Now you can derive the equation for Common Chord as :-

S1 - S2 = 0

(proof is given by S1 + u S2 = 0 i.e family of circles )

So

S1 - S2 = 0

(x^2 + y^2 + 2g1x + 2f1y + c1) -

(x^2 + y^2 + 2g2x + 2f2y + c2) = 0

2(g1-g2) x + 2(f1-f2) y + (c1-c2) = 0

Now let the equation for normal Normal to the Circle 1 is

ax + by + c = 0

Since it can be find by using the concept of family of straight lines i.e.

L1 + u L2 = 0

L1 = 2(g1-g2) x + 2(f1-f2) y + (c1-c2) = 0

L2 = ax + by + c = 0

So

2(g1-g2) x + 2(f1-f2) y + (c1-c2) +