II. 1. Find the equation of the common tangent of the following circles at their point of contact
(i) x2 + y2 + 10x – 2y + 22 = 0, x2 + y2 + 2x – 8y + 8 = 0
(ii) x2 + y2 - 8y - 4 = 0, x2 + y2 – 2x – 4y = 0
+
Answers
SOLUTION
TO DETERMINE
The equation of the common tangent of the following circles at their point of contact
(i) x² + y² + 10x – 2y + 22 = 0
x² + y² + 2x – 8y + 8 = 0
(ii) x² + y² - 8y - 4 = 0
x² + y² – 2x – 4y = 0
EVALUATION
(i) Here the given equation of the circles
x² + y² + 10x – 2y + 22 = 0
x² + y² + 2x – 8y + 8 = 0
The equation of the conic passing through the circles
x² + y² + 10x – 2y + 22 + k ( x² + y² + 2x – 8y + 8 ) = 0
⇒ (1 + k) (x² + y²) + ( 10 + 2k)x - ( 2+8k)y + (22+8k) = 0
Since the above equation represents equation of a line
So coefficient of x² = coefficient of y² = 0
∴ 1 + k = 0
⇒ k = - 1
So the required equation of tangent is obtained by replacing k by - 1
Hence equation of the common tangent of the circles at their point of contact
8x + 6y + 14 = 0
⇒ 4x + 3y + 7 = 0
(ii) Here the given equation of the circles
x² + y² - 8y - 4 = 0
x² + y² – 2x – 4y = 0
Hence equation of the common tangent of the circles at their point of contact
- 6x + 4y - 4 = 0
⇒ 3x - 2y + 2 = 0
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Step-by-step explanation:
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