It is given that we need to find the equation of the circle with centre (3, 4) and touches the straight line 5x + 12y – 1 = 0.
Since, circle with centre (3, 4) and having a radius 62/13.
As, the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
substituting the values in the equation, we get
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }3 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }4 \right)}^{2}}~=\text{ }{{\left( 62/13 \right)}^{2}} \\ {{x}^{2}}~-\text{ }6x\text{ }+\text{ }9\text{ }+\text{ }{{y}^{2}}~-\text{ }8y\text{ }+\text{ }16\text{ }=\text{ }3844/169 \\ 169{{x}^{2}}~+\text{ }169{{y}^{2}}~-\text{ }1014x\text{ }\text{ }1352y\text{ }+\text{ }4225\text{ }=\text{ }3844 \\ 169{{x}^{2}}~+\text{ }169{{y}^{2}}~-\text{ }1014x\text{ }\text{ }1352y\text{ }+\text{ }381\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ecb3989aef49e9a45e9e06007145c450_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is 169x2 + 169y2 – 1014x – 1352y + 381 = 0.