TRANSCRIBE THE SOLUTIONS IN DIGITAL FORMAT PLEASE we are to find the equation of a circleCenter at (7,-6) and that is tangent tothe line with equation23.-³7--19 = 0Here the given point is (7,-6)and the given tangent lineis22-37 -19=0perpendicular distance from (7₁-6) to the12(7)-3(-6)-19122-37-19 = 0 is√2+(-3)114 +18-19)ThelineSo equation of the circleor,oror1=x² + y²₁²-14m² +12.127(x-7)+ (4-6-6)) = (√13) ~√4+913√132x-37-19-01is:: Equation of the circle is,x² + y² = 14x+12y +72=0+72=0x²²-14x +49 + y ₁²+12 7+36 = 13in1²+ y² _14m +12 y +49 +36-13 = 0=√13(7,-6)2.The points A (-7,-4) and 13 (8,-1)circle.ends ofdiameter ofCenterRadiusor,=or,of( -748, -4++-1))2207,aor,orthe-5( 1121 )/2/2 )of√ (8-1) + (-1-(-3)*circle is= √ ( 152 ) ² + (-1 + 2/2 ) ²=√(-2) ¹3√√262Equation of the circletheacircle is.= 1/² √√225 +9= = √234 =-x+25tấty tổng= ₁ + y² +57.+4(x-1) ²+ (1-(²-3))² = (3-√2+²) ²√√26 2sty 45g -=sĩ tỷ ngn+=ax² + y² = x + 57:x² + y² = x + 5y-52=0A(-7,-4)= 52=234-1-25។1234-4-4234425are the06/12B3 (8,-1)

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We are to find the equation of a circle conter at (7₁-6) and that is tangent to the line with equation 2n-37-19 = 0 Here the given point is (7,-6) and the given tongent line is 22-37-19=0 is The perpendicular distance from (7₁-6) to the line 2m-37-19 = 0 12(7)-3(-6)-191 = = √2+(-3)+ 114+18-19) So equation of the circle 2x-3y-19:0 √4+9 13 (x-7)~ + (9-(-6))² = (√13) ~ - Equation of the circle is, x² + y²-14x+12y + 72 =0 is √13 or x²-14x +49 + y²₁+12 7+36 = 13 or n²+ y²-14m +12y +49 +36-13 = 0 or, n²+y^²-142 +12y + 72 = 0 (7₁-6) 2. ends of The points A (-7,-4) and B (8,-1) are the diameter of a circle. a Center of the circle is (-7+8, -4++)) Radim = (-1/2)/² ) circle is of the √ (8-12 ) ² + (-1-(-3)* = √( 1² )² + (-1 + ²) ² = √(4) + (-+-)* = 1/² √225+9 = 1/2 √234 - 3-√26 2 Equation of the circle or x²-x+ ² + y²+ 5y + ²4 234 or, styên t5g 2 or m² + g²-n + 5y or, or, (x-1) ²+ (a-(-))² = (3-√2²) ² = A(-2-4) 234-1-25 ។ x²+²x+57 = 52 x²+x+57-52 = 0 0 (315)) 234 4 B (8,-1)