Question

Choose the correct statement about two circles whose equations are given below : x² + y² - 10x - 10y + 41 = 0 ; x² + y² - 22x - 10y + 137 = 0

• A. circles have two meeting points
• B. circles have no meeting point
• C. circles have only one meeting point
• D. circles have same centre

Hint:

If $d = r_1 + r_2$, circles touch externally. If $d = |r_1 - r_2|$, they touch internally.

Answer 1

The Correct Option is C

Step 1: Find Center and Radius for $C_1$: $x^2 + y^2 - 10x - 10y + 41 = 0$. $C_1 = (5, 5)$, $r_1 = \sqrt{5^2 + 5^2 - 41} = \sqrt{25+25-41} = \sqrt{9} = 3$.
Step 2: Find Center and Radius for $C_2$: $x^2 + y^2 - 22x - 10y + 137 = 0$. $C_2 = (11, 5)$, $r_2 = \sqrt{11^2 + 5^2 - 137} = \sqrt{121+25-137} = \sqrt{9} = 3$.
Step 3: Distance between centers $d = \sqrt{(11-5)^2 + (5-5)^2} = \sqrt{6^2 + 0^2} = 6$.
Step 4: Check condition: $r_1 + r_2 = 3 + 3 = 6$.
Step 5: Since $d = r_1 + r_2$, the circles touch each other externally at exactly one point. [Image of two circles touching externally]