3,000 Solved Problems in Calculus 28 - CIRCLES 21 4.18 Find the standard equation of a circle with radius 13 that passes through the origin and whose

# 3,000 Solved Problems in Calculus 28 - CIRCLES 21 4.18 Find...

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CIRCLES Find the standard equation of a circle with radius 13 that passes through the origin, and whose center has abscissa -12. Let the center be (-12, ft). The distance formula yields 144+ft 2 = 169, b 2 = 25, and b = ±5. Hence, there are two circles, with equations (x + 12) 2 + (y - 5) 2 = 169 and (jc + 12) 2 + (y + 5) 2 = 169. Find the standard equation of the circle with center at (1, 3) and tangent to the line 5x - I2y -8 = 0. 4.20 4.21 The radius is the perpendicular distance from the center (1,3) to the line: standard equation is (x - I) 2 + ( y - 3) 2 = 9. Find the standard equation of the circle passing through (-2, 1) and tangent to the line 3x - 2y - 6 at the point (4,3). Since the circle passes through (-2, 1) and (4, 3), its center (a, b) is on the perpendicular bisector of the segment connecting those points. The center must also be on the line perpendicular to 3x-2y = 6 at (4, 3). The equation of the perpendicular bisector of the segment is found to be 3x + y = 5. The equation of the line perpendicular to 3* - 2y = 6 at (4, 3) turns out to be 2x + 3y = 17. Solving 3* + y = 5 and 2* + 3.y = 17 simultaneously, we find x = -\$ and y = % . Then the radius the required equation is (x + f ) 2 + ( y - ^ ) 2 = ^ . Find a formula for the length / of the tangent from an exterior point P(x l ,y l ) to the circle (x - a) 2 + ( y-b)2 = r2. See Fig. 4-2.