The locus of the centre of a circle which touches externally the circle
Answers
Answered by
0
Answer:
- let the equation of the given circle is (x-a)²+(y-b)²=R²
- and the externally touches circle's equation is (x-h)²+(y-k)²=r²
- because both circle touches externally son the sum of their radius is equal to distance between their center.
- so R+r=√[(a-h)²+(b-k)²]
- by squaring both side
- (r+R)²=(h-a)²+(k-b)²
- or (h-a)²+(k-b)²=(r+R)²
- for finding the locus of the center we put h=x and k=y
- by putting these values we found
- (x-a)²+(y-b)²=(r+R)²
- so the locus of center is also a center whose center is at known center and radius is equal to the sum of the both radius.
Step-by-step explanation:
Similar questions