Find the equations of the circles which touch the axis of x at a distance of 4 from the origin and cut off an intercept of 6 from the axis of y
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Answer -
Given:
PQ = 6 units
OM = 4 units
QN = PN = 3 Units
CN = MO = 4 Units
CQ² = QN² + NC²
CQ = √ 3² + 4²
= 5 units
Now CQ = CM = radius = 5 units
co - ordinate of C = (5,4)
and radius = 5 units so the equation of circle would be :
(x - 5)² + (y-4)² = 5²
x² + y² - 10x - 8y + 41 - 25 = 0
x² + y² - 10x - 8y + 16 = 0
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Given:
PQ = 6 units
OM = 4 units
QN = PN = 3 Units
CN = MO = 4 Units
CQ² = QN² + NC²
CQ = √ 3² + 4²
= 5 units
Now CQ = CM = radius = 5 units
co - ordinate of C = (5,4)
and radius = 5 units so the equation of circle would be :
(x - 5)² + (y-4)² = 5²
x² + y² - 10x - 8y + 41 - 25 = 0
x² + y² - 10x - 8y + 16 = 0
Plz mark as Brainliest
Happy to help
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