Question

Equation of circle with centre (4,3 and touching the line 5x-12y-10=0

Answer 1

Answer:

x²+y²-8x-6x+21=0

Step-by-step explanation:

given, the centre of the circle is (4,3)

and touches the line 5x-12y-10=0

=> the radius a = | 5(4)-12(3)-10 |  / √(5²+(-12)²)

                         = | 20-36-10 | / √(25+144)

                         = | -26 | / √169

                         =26 / 13 = 2 units

so, the radius, a = 2 units

therefore the equation of the circle can be written as,

(x-h)²+(y-k)²=a²

(x-4)²+(y-3)² = 2²

x²-8x+16+y²-6x+9=4

x²+y²-8x-6x+25-4=0

x²+y²-8x-6x+21=0