Question

The centre of a circle is (4a-2, 6a+2) and is passing through the point (-6, -2) if the diameter of the circle is 40 then find the value of a

Answer 1

Step-by-step explanation:

Given The centre of a circle is (4a-2, 6a+2) and is passing through the point (-6, -2) if the diameter of the circle is 40 then find the value of a

• Given diameter = 40 units
• So radius of circle = 20 units
• We know that distance = √(x2 – x1)^2 + (y2 – y1)^2
•                                    √[(4a – 2 – (- 6)]^2 + [(6a + 2 – (- 2)]^2 = 20
•                                   √[4a + 4]^2 + [6a + 4]^2 = 20
•                               √(16a^2 + 16 + 32a) + (36a^2 + 16 + 48a)  = 20
•                                √52a^2 + 80a + 32 = 20
• Squaring both sides we get
•                              52a^2 + 80a + 32 = 400
•                              52a^2 + 80a = 368
•                                52a^2 + 80a – 368 = 0
•                         So 13a^2 + 20a – 92 = 0 (taking 4 common )
•                              13a^2 + 46a – 26a – 92 = 0
•                     So    a(13a + 46) – 2(13a + 46) = 0
•                           (13a + 46)(a – 2) = 0
•                    So a = 2, - 46 / 13
•                     Or a = 2 since distance cannot be negative.

Reference link will be

https://brainly.in/question/14363718

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