Pq is a chord of length …

Given radius, OP = OQ = 5 cm
Length of chord, PQ = 4 cm
OT {tex}\perp{/tex} PQ,
{tex}\therefore{/tex} PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right {tex}\triangle{/tex}OPM, OP² = PM² + OM²
{tex}\Rightarrow{/tex} 5² = 4² + OM²
{tex}\Rightarrow{/tex} OM² = 25 – 16 = 9
Hence OM = 3cm
In right {tex}\triangle{/tex}PTM, PT² = TM² + PM² {tex}\rightarrow{/tex} (1)
OPT = 90^o [Radius is perpendicular to tangent at point of contact]
In right {tex}\triangle{/tex}OPT, OT² = PT² + OP² {tex}\rightarrow{/tex} (2)
From equations (1) and (2), we get
OT² = (TM² + PM²) + OP²
{tex}\Rightarrow{/tex} (TM + OM)² = (TM² + PM²) + OP²
{tex}\Rightarrow{/tex} TM² + OM² + 2 {tex}\times{/tex} TM {tex}\times{/tex} OM = TM² + PM² + OP²
{tex}\Rightarrow{/tex} OM² + 2 {tex}\times{/tex} TM {tex}\times{/tex} OM = PM² + OP²
{tex}\Rightarrow{/tex} 3² + 2 {tex}\times{/tex} TM {tex}\times{/tex} 3 = 4² + 5²
{tex}\Rightarrow{/tex} 9 + 6TM = 16 + 25
{tex}\Rightarrow{/tex} 6TM = 32
{tex}\Rightarrow{/tex} TM ={tex}\frac{16}3{/tex}
Equation (1) becomes,
PT² = TM² + PM²
= ({tex}\frac{16}3{/tex})² + 4²
= ({tex}\frac{256}9{/tex}) + 16 = {tex}\frac{256+ 144}9{/tex}
= ({tex}\frac{400}9{/tex}) = ({tex}\frac{20}3{/tex})²
Hence PT ={tex}\frac{20}3{/tex}
Thus, the length of tangent PT is ({tex}\frac{20}3{/tex}) cm.