find the equation of circle having its centre at the point of intersection of 2x-3y+4=0 and 3x+4y-5=0 and passing through the origin
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ANSWER
Intersection point of 2x−3y+4=0 and 3x+4y−5=0-
2x−3y+4=0
Multiplying above equation by 4, we get
8x−12y+16=0.....(1)
3x+4y−5=0
Multiplying above equation by 3, we get
9x+12y−15=0.....(2)
Adding eq
n
(1)&(2), we have
(4x−12y+16)+(9x+12y−15)=0
⇒13x=−1
⇒x=
17
−1
Substituting the value of x in eq
n
(1), we have
17
−4
−12y+16=0
⇒12y=
17
264
⇒y=
17
22
Hence the centre of circle will be (
17
−1
,
17
22
).
As the circle touches the origin, hence the distance between the centre of circle to the origin will be the radius of circle.
∴r=
(
17
−1
−0)
2
+(
17
22
−0)
2
⇒r=
289
1
+
289
484
⇒r
2
=
289
485
As we know that equation of a circle whose centre at (h,k) and radius r is given by-
(x−h)
2
+(y−k)
2
=r
2
Therefore, equation of the given circle will be-
(x−(
17
−1
))
2
+(y−
17
22
)
2
=
289
485
⇒(x+
17
1
)
2
+(y−
17
22
)
2
=
289
485
⇒(17x+1)
2
+(17y−22)
2
=485
Hence the required answer is (17x+1)
2
+(17y−22)
2
=485.
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