Question

the eq of chord of contact of the circle x^2+y^2+4x+6y-12=0 with respect to the point(2,3)

Answer 1

Solution :

x2+y2−4x+6y−12=0x2+y2−4x+6y−12=0

Centre (2, -3)

Radius 4+9+12−−−−−−−−√=54+9+12=5

Distance b/w two centres c1(2,−3)c1(2,−3)and C2(−3,1)C2(−3,1)

d=(2+3)2+(−3−2)2−−−−−−−−−−−−−−−−√=50−−√d=(2+3)2+(−3−2)2=50

Radius of (S)=52+(50−−√)2−−−−−−−−−−√=75−−√=53–√(S)=52+(50)2=75=53

Answer 2

Hola User_______________

Here is Your Answer..!!
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↪Actually welcome to the concept of the CIRCLES...

↪Basically we know that ..

↪CHORD OF CONTACT IS GIVEN BY

↪T=0

↪such that equation of tangent = 0

↪therefore

↪xx1 + yy1 + g (x+x1) + f (y+y1) + c =0
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