a circle with a centre of (0,0) passes thought the point (-3,4) what is the other point of the circle
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Answers
Step-by-step explanation:
Limiting Point of system of co-axial circles are the centers of the point circles belonging to the family.
As we know , every circle passing through limiting points of a coaxial system is orthogonal to all circles of the system.
So, any circle passing through a point and orthogonal to a given circle, will have its center on the radical axis of that point circle and that given circle. (or of the coaxial system)
So, let there be a circle with radius 0 and passes through (3,4). Equation of this circle S
1
:(x−3)
2
+(y−4)
2
=0
2
Given circle S
2
:x
2
+y
2
−a
2
=0
∴ Equation of Radical Axis :S
1
−S
2
=0
−6x−8y+25+a
2
=0
This is the equation of a radical axis of a co-axial system of circles of which (3,4) is a limiting point. Hence, this will be the required locus.
Distance of (0,0) from this Radical Axis is 25. So,
36+64
∣
∣
∣
25+a
2
∣
∣
∣
=25
⇒25+a
2
=250
∴a
2
=225
Step-by-step explanation:
Limiting Point of system of co-axial circles are the centers of the point circles belonging to the family.
As we know , every circle passing through limiting points of a coaxial system is orthogonal to all circles of the system.
So, any circle passing through a point and orthogonal to a given circle, will have its center on the radical axis of that point circle and that given circle. (or of the coaxial system)
So, let there be a circle with radius 0 and passes through (3,4). Equation of this circle S
1
:(x−3)
2
+(y−4)
2
=0
2
Given circle S
2
:x
2
+y
2
−a
2
=0
∴ Equation of Radical Axis :S
1
−S
2
=0
−6x−8y+25+a
2
=0
This is the equation of a radical axis of a co-axial system of circles of which (3,4) is a limiting point. Hence, this will be the required locus.
Distance of (0,0) from this Radical Axis is 25. So,
36+64
∣
∣
∣
25+a
2
∣
∣
∣
=25
⇒25+a
2
=250
∴a
2
=225