prove that two circles passaes through only two point's
Answers
Answer:
Let the equation of the circle be
x^{2} + y^{2} + 2gx + 2f\ y + \lambda = 0x
2
+y
2
+2gx+2f y+λ=0.
Since it passes through (0, a)(0,a) and (0, -a)(0,−a).
\therefore a^{2} = 2f a + \lambda = 0∴a
2
=2fa+λ=0
a^{2} - 2fa + \lambda = 0a
2
−2fa+λ=0.
Subtracting, we get 4fa = 0, \therefore f = 04fa=0,∴f=0; and \lambda =-a^{2}λ=−a
2
.
Hence the circle becomes
x^{2} + y^{2} + 2gx - a^{2} = 0x
2
+y
2
+2gx−a
2
=0.
Centre is (-g, 0)(−g,0)
and radius = \sqrt {g^{2} - \lambda} = \sqrt {g^{2} + a^{2}}=
g
2
−λ
=
g
2
+a
2
It is given that the circle touches the line
y = mx + cy=mx+c or mx - y + c = 0mx−y+c=0.
Hence perpendicular from centre should be equal to radius
\dfrac {-mg + c}{\sqrt {(m^{2} + 1)}} = \sqrt {g^{2} + a^{2}}
(m
2
+1)
−mg+c
=
g
2
+a
2
.
Square (c - mg)^{2} = (g^{2} + a^{2})(m^{2} + 1)(c−mg)
2
=(g
2
+a
2
)(m
2
+1)
or g^{2} + 2gmc + \left \{a^{2} (1 + m^{2}) - c^{2}\right \} = 0g
2
+2gmc+{a
2
(1+m
2
)−c
2
}=0.
\therefore g_{1}g_{2} = a^{2}(1 + m^{2}) - c^{2} ..... (1)∴g
1
g
2
=a
2
(1+m
2
)−c
2
.....(1)
Above is a quadratic in gg and gives us two values of gg showing that there will be two circles which satisfy the given conditions. If the two values of gg be g_{1}g
1
and g_{2}g
2
then the two circles are
x^{2} + y^{2} + 2g_{1} x - a^{2} = 0x
2
+y
2
+2g
1
x−a
2
=0
and x^{2} + y^{2} + 2g_{2}x - a^{2} = 0x
2
+y
2
+2g
2
x−a
2
=0.
These two will intersect orthogonally if
2g_{1}g_{2} = 2f_{1}f_{2} = c_{1} + c_{2}2g
1
g
2
=2f
1
f
2
=c
1
+c
2
2\left \{a^{2}(1 + m^{2}) - c^{2}\right \} + 0 = -a^{2} - a^{2} = -2a^{2}2{a
2
(1+m
2
)−c
2
}+0=−a
2
−a
2
=−2a
2
or a^{2} (1 + m^{2}) - c^{2} = -a^{2}a
2
(1+m
2
)−c
2
=−a
2
or a^{2} (2 + m^{2}) = c^{2}a
2
(2+m
2
)=c
2
is the required condition
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