Circles $$\(C_{1}\)$$ and $$\(C_{2}\)$$ have equations $$\[ x^{2}+y^{2}+6 x-10 y+18=0 \text { and }(x-9)^{2}+(y+4)^{2}-64=0 \]$$ respectively. Find the distance between the centres of the circles. ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . ........................................................................................................................................................ . $$\(P\)$$ and $$\(Q\)$$ are points on $$\(C_{1}\)$$ and $$\(C_{2}\)$$ respectively. The distance between $$\(P\)$$ and $$\(Q\)$$ is denoted by $$\(d\)$$.
Exam No:9709_w24_qp_11 Year:2024 Question No:6(a)
Answer:

Knowledge points:
1.3.3 understand that the equation represents the circle with centre and radius (Including use of the expanded form.)
Solution:
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