MHT CET · Maths · Circle
Given two circles \(\quad x^2+y^2+8 x-6 y-24=0 \quad\) and \(x^2+y^2-4 x+10 y+20=0\). Then they are
- A Disjoint.
- B Concentric,
- C Touching internally
- D Touching externally.
Answer & Solution
Correct Answer
(D) Touching externally.
Step-by-step Solution
Detailed explanation
\( x^2+y^2+8 x-6 y-24=0, C_1 \equiv(-4,3), r_1=7 \)
\( x^2+y^2-4 x+10 y+20=0, C_2 \equiv(2,-5), r_2=3 \)
\( C_1 C_2=\sqrt{(2+4)^2+(-5-3)^2}=10 \text { and } r_1+r_2=7\)\(+3=10\)
\(\because C_1 C_2=r_1+r_2\) hence, the tow circles touching externally
\( x^2+y^2-4 x+10 y+20=0, C_2 \equiv(2,-5), r_2=3 \)
\( C_1 C_2=\sqrt{(2+4)^2+(-5-3)^2}=10 \text { and } r_1+r_2=7\)\(+3=10\)
\(\because C_1 C_2=r_1+r_2\) hence, the tow circles touching externally
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