solve it frds fast
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Q.1 Given the equation of the circle is x
the equation of the circle is x 2
the equation of the circle is x 2 +y
the equation of the circle is x 2 +y 2
the equation of the circle is x 2 +y 2 −6x−8y+24=0
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0or, (x−3)
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0or, (x−3) 2
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0or, (x−3) 2 +(y−4)
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0or, (x−3) 2 +(y−4) 2
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0or, (x−3) 2 +(y−4) 2 =1
the equation of the circle is x 2 +y 2 −6x−8y+24=0or, (x 2 −6x+9)+(y 2 −8y+16)−1=0or, (x−3) 2 +(y−4) 2 =1So, the centre is (3,4) and radius is 1.
Q2.Let point P be (x,y)
Given A(1,2),B(2,−3),C(−2,3) and
PA
2
+PB
2
=2PC
2
⟹(
(x−1)
2
+(y−2)
2
)
2
+(
(x−2)
2
+(y+3)
2
)
2
=2(
(x+2)
2
+(y−3)
2
)
2
⟹x
2
+1−2x+y
2
+4−4y+x
2
+4−4x+y
2
+9+6y=2(x
2
+4+4x+y
2
+9−6y)
⟹2x
2
+2y
2
−6x+2y+18=2x
2
+2y
2
+8x−12y+26
⟹14x−14y+8=0
⟹7x−7y+4=0
Therefore, the locus of point P is 7x−7y+4=0