find the equation of the circle in the first quadrant which touches both the axes and whose center lies on the straight line ax + by + c = 0
Answers
Answered by
0
Answer:
Step-by-step explanation:
Given,
⇒ circle of radius a touches both axis in 1st quadrant so its centre will be (a,a).
∴ Required equation
⇒(x−a)
2
+(y−a)
2
=a
2
⇒x
2
+a
2
−2ax+y
2
+a
2
−2ay=a
2
⇒x
2
+y
2
−2ax−2ay+2a
2
−a
2
=0
⇒x
2
−y
2
−2ax−2ay+a
2
=0.
Answered by
1
Answer:
Given,
⇒ circle of radius a touches both axis in 1st quadrant so its centre will be (a,a).
∴ Required equation
⇒(x−a)
2
+(y−a)
2
=a
2
⇒x
2
+a
2
−2ax+y
2
+a
2
−2ay=a
2
⇒x
2
+y
2
−2ax−2ay+2a
2
−a
2
=0
⇒x
2
−y
2
−2ax−2ay+a
2
=0.
Step-by-step explanation:
Similar questions